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feat: add solutions to lc problem: No.1975
No.1975.Maximum Matrix Sum
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‎solution/1900-1999/1975.Maximum Matrix Sum/README.md

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@@ -49,22 +49,143 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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如果矩阵中存在零，或者矩阵中负数的个数为偶数，那么最大和就是矩阵中所有元素的绝对值之和。
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否则，说明矩阵中有奇数个负数，最终一定会剩下一个负数，我们选择绝对值最小的数，将其变为负数，这样可以使得最终的和最大。
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时间复杂度 $O(m\times n)$，其中 $m$ 和 $n$ 分别是矩阵的行数和列数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxMatrixSum(self, matrix: List[List[int]]) -> int:
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s = cnt = 0
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mi = inf
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for row in matrix:
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for v in row:
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s += abs(v)
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mi = min(mi, abs(v))
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if v < 0:
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cnt += 1
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if cnt % 2 == 0 or mi == 0:
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return s
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return s - mi * 2
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public long maxMatrixSum(int[][] matrix) {
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long s = 0;
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int cnt = 0;
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int mi = Integer.MAX_VALUE;
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for (var row : matrix) {
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for (var v : row) {
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s += Math.abs(v);
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mi = Math.min(mi, Math.abs(v));
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if (v < 0) {
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++cnt;
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}
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}
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}
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if (cnt % 2 == 0 || mi == 0) {
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return s;
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}
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return s - mi * 2;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maxMatrixSum(vector<vector<int>>& matrix) {
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long long s = 0;
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int cnt = 0, mi = INT_MAX;
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for (auto& row : matrix) {
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for (int& v : row) {
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s += abs(v);
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mi = min(mi, abs(v));
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cnt += v < 0;
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}
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}
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if (cnt % 2 == 0 || mi == 0) return s;
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return s - mi * 2;
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}
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};
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```
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### **Go**
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```go
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func maxMatrixSum(matrix [][]int) int64 {
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s := 0
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cnt, mi := 0, math.MaxInt32
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for _, row := range matrix {
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for _, v := range row {
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s += abs(v)
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mi = min(mi, abs(v))
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if v < 0 {
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cnt++
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}
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}
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}
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if cnt%2 == 1 {
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s -= mi * 2
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}
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return int64(s)
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[][]} matrix
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* @return {number}
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*/
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var maxMatrixSum = function (matrix) {
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let cnt = 0;
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let s = 0;
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let mi = Infinity;
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for (const row of matrix) {
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for (const v of row) {
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s += Math.abs(v);
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mi = Math.min(mi, Math.abs(v));
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cnt += v < 0;
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}
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}
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if (cnt % 2 == 0) {
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return s;
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}
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return s - mi * 2;
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};
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```
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### **...**

‎solution/1900-1999/1975.Maximum Matrix Sum/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maxMatrixSum(self, matrix: List[List[int]]) -> int:
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s = cnt = 0
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mi = inf
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for row in matrix:
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for v in row:
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s += abs(v)
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mi = min(mi, abs(v))
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if v < 0:
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cnt += 1
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if cnt % 2 == 0 or mi == 0:
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return s
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return s - mi * 2
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```
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### **Java**
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```java
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class Solution {
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public long maxMatrixSum(int[][] matrix) {
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long s = 0;
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int cnt = 0;
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int mi = Integer.MAX_VALUE;
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for (var row : matrix) {
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for (var v : row) {
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s += Math.abs(v);
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mi = Math.min(mi, Math.abs(v));
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if (v < 0) {
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++cnt;
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}
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}
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}
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if (cnt % 2 == 0 || mi == 0) {
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return s;
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}
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return s - mi * 2;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maxMatrixSum(vector<vector<int>>& matrix) {
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long long s = 0;
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int cnt = 0, mi = INT_MAX;
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for (auto& row : matrix) {
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for (int& v : row) {
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s += abs(v);
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mi = min(mi, abs(v));
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cnt += v < 0;
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}
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}
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if (cnt % 2 == 0 || mi == 0) return s;
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return s - mi * 2;
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}
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};
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```
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### **Go**
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```go
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func maxMatrixSum(matrix [][]int) int64 {
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s := 0
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cnt, mi := 0, math.MaxInt32
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for _, row := range matrix {
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for _, v := range row {
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s += abs(v)
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mi = min(mi, abs(v))
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if v < 0 {
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cnt++
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}
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}
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}
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if cnt%2 == 1 {
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s -= mi * 2
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}
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return int64(s)
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[][]} matrix
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* @return {number}
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*/
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var maxMatrixSum = function (matrix) {
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let cnt = 0;
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let s = 0;
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let mi = Infinity;
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for (const row of matrix) {
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for (const v of row) {
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s += Math.abs(v);
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mi = Math.min(mi, Math.abs(v));
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cnt += v < 0;
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}
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}
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if (cnt % 2 == 0) {
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return s;
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}
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return s - mi * 2;
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};
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```
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### **...**

‎solution/1900-1999/1975.Maximum Matrix Sum/Solution.cpp

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@@ -0,0 +1,16 @@
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class Solution {
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public:
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long long maxMatrixSum(vector<vector<int>>& matrix) {
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long long s = 0;
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int cnt = 0, mi = INT_MAX;
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for (auto& row : matrix) {
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for (int& v : row) {
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s += abs(v);
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mi = min(mi, abs(v));
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cnt += v < 0;
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}
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}
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if (cnt % 2 == 0 || mi == 0) return s;
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return s - mi * 2;
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}
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};

‎solution/1900-1999/1975.Maximum Matrix Sum/Solution.go

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func maxMatrixSum(matrix [][]int) int64 {
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s := 0
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cnt, mi := 0, math.MaxInt32
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for _, row := range matrix {
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for _, v := range row {
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s += abs(v)
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mi = min(mi, abs(v))
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if v < 0 {
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cnt++
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}
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}
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}
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if cnt%2 == 1 {
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s -= mi * 2
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}
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return int64(s)
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}

‎solution/1900-1999/1975.Maximum Matrix Sum/Solution.java

Lines changed: 20 additions & 0 deletions
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class Solution {
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public long maxMatrixSum(int[][] matrix) {
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long s = 0;
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int cnt = 0;
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int mi = Integer.MAX_VALUE;
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for (var row : matrix) {
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for (var v : row) {
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s += Math.abs(v);
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mi = Math.min(mi, Math.abs(v));
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if (v < 0) {
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++cnt;
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}
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}
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}
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if (cnt % 2 == 0 || mi == 0) {
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return s;
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}
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return s - mi * 2;
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}
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}

‎solution/1900-1999/1975.Maximum Matrix Sum/Solution.js

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@@ -0,0 +1,20 @@
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/**
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* @param {number[][]} matrix
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* @return {number}
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*/
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var maxMatrixSum = function (matrix) {
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let cnt = 0;
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let s = 0;
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let mi = Infinity;
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for (const row of matrix) {
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for (const v of row) {
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s += Math.abs(v);
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mi = Math.min(mi, Math.abs(v));
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cnt += v < 0;
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}
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}
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if (cnt % 2 == 0) {
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return s;
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}
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return s - mi * 2;
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};

‎solution/1900-1999/1975.Maximum Matrix Sum/Solution.py

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class Solution:
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def maxMatrixSum(self, matrix: List[List[int]]) -> int:
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s = cnt = 0
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mi = inf
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for row in matrix:
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for v in row:
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s += abs(v)
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mi = min(mi, abs(v))
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if v < 0:
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cnt += 1
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if cnt % 2 == 0 or mi == 0:
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return s
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return s - mi * 2

‎solution/2400-2499/2416.Sum of Prefix Scores of Strings/README.md

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用前缀树维护所有字符串的前缀以及每个前缀出现的次数。
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然后遍历每个字符串，累加每个前缀的出现次数即可。
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时间复杂度 $O(n\times m)$。其中 $n$, $m$ 分别为字符串数组 `words` 的长度和其中字符串的最大长度。
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<!-- tabs:start -->
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### **Python3**

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