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feat: add solutions to lc problem: No.1278
No.1278.Palindrome Partitioning III
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‎solution/1200-1299/1278.Palindrome Partitioning III/README.md

Lines changed: 133 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -51,22 +51,154 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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定义 $dp[i][j]$ 表示将字符串 $s$ 的前 $i$ 个字符分割成 $j$ 个回文串所需要的最少修改次数，我们假定 $i$ 下标从 $1$ 开始，答案为 $dp[n][k]$。
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对于 $dp[i][j]$，我们可以枚举第 $j-1$ 个回文串的最后一个字符的位置 $h$，那么 $dp[i][j]$ 就等于 $dp[h][j-1] + g[h][i-1]$ 的较小值，其中 $g[h][i-1]$ 表示将字符串 $s[h..i-1]$ 变成回文串所需要的最少修改次数（这一部分我们可以通过预处理得到，时间复杂度 $O(n^2)$。
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时间复杂度 $O(n^2\times k)$。其中 $n$ 为字符串 $s$ 的长度。
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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
61-
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class Solution:
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def palindromePartition(self, s: str, k: int) -> int:
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n = len(s)
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g = [[0] * n for _ in range(n)]
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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g[i][j] = int(s[i] != s[j])
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if i + 1 < j:
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g[i][j] += g[i + 1][j - 1]
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f = [[0] * (k + 1) for _ in range(n + 1)]
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for i in range(1, n + 1):
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for j in range(1, min(i, k) + 1):
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if j == 1:
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f[i][j] = g[0][i - 1]
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else:
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f[i][j] = inf
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for h in range(j - 1, i):
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f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1])
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return f[n][k]
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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 int palindromePartition(String s, int k) {
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int n = s.length();
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int[][] g = new int[n][n];
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for (int i = n - 1; i >= 0; --i) {
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for (int j = i; j < n; ++j) {
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g[i][j] = s.charAt(i) != s.charAt(j) ? 1 : 0;
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if (i + 1 < j) {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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int[][] f = new int[n + 1][k + 1];
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j <= Math.min(i, k); ++j) {
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if (j == 1) {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = 10000;
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for (int h = j - 1; h < i; ++h) {
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f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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return f[n][k];
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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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int palindromePartition(string s, int k) {
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int n = s.size();
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vector<vector<int>> g(n, vector<int>(n));
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for (int i = n - 1; i >= 0; --i) {
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for (int j = i; j < n; ++j) {
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g[i][j] = s[i] != s[j] ? 1 : 0;
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if (i + 1 < j) g[i][j] += g[i + 1][j - 1];
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}
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}
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vector<vector<int>> f(n + 1, vector<int>(k + 1));
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j <= min(i, k); ++j) {
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if (j == 1) {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = 10000;
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for (int h = j - 1; h < i; ++h) {
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f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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return f[n][k];
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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 palindromePartition(s string, k int) int {
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n := len(s)
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g := make([][]int, n)
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for i := range g {
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g[i] = make([]int, n)
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}
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for i := n - 1; i >= 0; i-- {
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for j := 1; j < n; j++ {
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if s[i] != s[j] {
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g[i][j] = 1
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}
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if i+1 < j {
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g[i][j] += g[i+1][j-1]
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}
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}
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}
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f := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, k+1)
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}
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for i := 1; i <= n; i++ {
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for j := 1; j <= min(i, k); j++ {
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if j == 1 {
184+
f[i][j] = g[0][i-1]
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} else {
186+
f[i][j] = 100000
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for h := j - 1; h < i; h++ {
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f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
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}
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}
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}
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}
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return f[n][k]
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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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```
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### **...**

‎solution/1200-1299/1278.Palindrome Partitioning III/README_EN.md

Lines changed: 125 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -51,13 +51,137 @@
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### **Python3**
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```python
54-
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class Solution:
55+
def palindromePartition(self, s: str, k: int) -> int:
56+
n = len(s)
57+
g = [[0] * n for _ in range(n)]
58+
for i in range(n - 1, -1, -1):
59+
for j in range(i + 1, n):
60+
g[i][j] = int(s[i] != s[j])
61+
if i + 1 < j:
62+
g[i][j] += g[i + 1][j - 1]
63+
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f = [[0] * (k + 1) for _ in range(n + 1)]
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for i in range(1, n + 1):
66+
for j in range(1, min(i, k) + 1):
67+
if j == 1:
68+
f[i][j] = g[0][i - 1]
69+
else:
70+
f[i][j] = inf
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for h in range(j - 1, i):
72+
f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1])
73+
return f[n][k]
5574
```
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### **Java**
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```java
79+
class Solution {
80+
public int palindromePartition(String s, int k) {
81+
int n = s.length();
82+
int[][] g = new int[n][n];
83+
for (int i = n - 1; i >= 0; --i) {
84+
for (int j = i; j < n; ++j) {
85+
g[i][j] = s.charAt(i) != s.charAt(j) ? 1 : 0;
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if (i + 1 < j) {
87+
g[i][j] += g[i + 1][j - 1];
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}
89+
}
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}
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int[][] f = new int[n + 1][k + 1];
92+
for (int i = 1; i <= n; ++i) {
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for (int j = 1; j <= Math.min(i, k); ++j) {
94+
if (j == 1) {
95+
f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = 10000;
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for (int h = j - 1; h < i; ++h) {
99+
f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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return f[n][k];
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}
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}
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```
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### **C++**
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```cpp
112+
class Solution {
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public:
114+
int palindromePartition(string s, int k) {
115+
int n = s.size();
116+
vector<vector<int>> g(n, vector<int>(n));
117+
for (int i = n - 1; i >= 0; --i) {
118+
for (int j = i; j < n; ++j) {
119+
g[i][j] = s[i] != s[j] ? 1 : 0;
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if (i + 1 < j) g[i][j] += g[i + 1][j - 1];
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}
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}
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vector<vector<int>> f(n + 1, vector<int>(k + 1));
124+
for (int i = 1; i <= n; ++i) {
125+
for (int j = 1; j <= min(i, k); ++j) {
126+
if (j == 1) {
127+
f[i][j] = g[0][i - 1];
128+
} else {
129+
f[i][j] = 10000;
130+
for (int h = j - 1; h < i; ++h) {
131+
f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
135+
}
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return f[n][k];
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}
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};
139+
```
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### **Go**
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```go
144+
func palindromePartition(s string, k int) int {
145+
n := len(s)
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g := make([][]int, n)
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for i := range g {
148+
g[i] = make([]int, n)
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}
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for i := n - 1; i >= 0; i-- {
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for j := 1; j < n; j++ {
152+
if s[i] != s[j] {
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g[i][j] = 1
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}
155+
if i+1 < j {
156+
g[i][j] += g[i+1][j-1]
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}
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}
159+
}
160+
f := make([][]int, n+1)
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for i := range f {
162+
f[i] = make([]int, k+1)
163+
}
164+
for i := 1; i <= n; i++ {
165+
for j := 1; j <= min(i, k); j++ {
166+
if j == 1 {
167+
f[i][j] = g[0][i-1]
168+
} else {
169+
f[i][j] = 100000
170+
for h := j - 1; h < i; h++ {
171+
f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
172+
}
173+
}
174+
}
175+
}
176+
return f[n][k]
177+
}
178+
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func min(a, b int) int {
180+
if a < b {
181+
return a
182+
}
183+
return b
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}
61185
```
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63187
### **...**

‎solution/1200-1299/1278.Palindrome Partitioning III/Solution.cpp

Lines changed: 17 additions & 22 deletions
Original file line numberDiff line numberDiff line change
@@ -1,32 +1,27 @@
11
class Solution {
22
public:
33
int palindromePartition(string s, int k) {
4-
int len = s.length();
5-
// cost[i][j] = min #changes needed to turn (s[i],...,s[i+j]) into a palindrome
6-
vector<vector<int>> cost(len, vector<int>(len));
7-
for (int i = 0; i < len; i++) {
8-
for (int j = i; j < len; j++) {
9-
int begin = i, end = j, cnt = 0;
10-
while (begin < end) {
11-
if (s[begin] != s[end])
12-
cnt++;
13-
begin++, end--;
14-
}
15-
cost[i][j - i] = cnt;
4+
int n = s.size();
5+
vector<vector<int>> g(n, vector<int>(n));
6+
for (int i = n - 1; i >= 0; --i) {
7+
for (int j = i; j < n; ++j) {
8+
g[i][j] = s[i] != s[j] ? 1 : 0;
9+
if (i + 1 < j) g[i][j] += g[i + 1][j - 1];
1610
}
1711
}
18-
// dp[i][j] = min #changes needed to split (s[i],...,s[len]) into j+1 palindromes
19-
vector<vector<int>> dp(len, vector<int>(k, INT_MAX));
20-
for (int i = 0; i < len; i++) {
21-
dp[i][0] = cost[i][len - 1 - i];
22-
}
23-
for (int kk = 1; kk < k; kk++) {
24-
for (int i = 0; i < len; i++) {
25-
for (int j = i + 1; j + kk - 1 < len; j++) {
26-
dp[i][kk] = min(dp[i][kk], dp[j][kk - 1] + cost[i][j - i - 1]);
12+
vector<vector<int>> f(n + 1, vector<int>(k + 1));
13+
for (int i = 1; i <= n; ++i) {
14+
for (int j = 1; j <= min(i, k); ++j) {
15+
if (j == 1) {
16+
f[i][j] = g[0][i - 1];
17+
} else {
18+
f[i][j] = 10000;
19+
for (int h = j - 1; h < i; ++h) {
20+
f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1]);
21+
}
2722
}
2823
}
2924
}
30-
return dp[0][k - 1];
25+
return f[n][k];
3126
}
3227
};

‎solution/1200-1299/1278.Palindrome Partitioning III/Solution.go

Lines changed: 41 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,41 @@
1+
func palindromePartition(s string, k int) int {
2+
n := len(s)
3+
g := make([][]int, n)
4+
for i := range g {
5+
g[i] = make([]int, n)
6+
}
7+
for i := n - 1; i >= 0; i-- {
8+
for j := 1; j < n; j++ {
9+
if s[i] != s[j] {
10+
g[i][j] = 1
11+
}
12+
if i+1 < j {
13+
g[i][j] += g[i+1][j-1]
14+
}
15+
}
16+
}
17+
f := make([][]int, n+1)
18+
for i := range f {
19+
f[i] = make([]int, k+1)
20+
}
21+
for i := 1; i <= n; i++ {
22+
for j := 1; j <= min(i, k); j++ {
23+
if j == 1 {
24+
f[i][j] = g[0][i-1]
25+
} else {
26+
f[i][j] = 100000
27+
for h := j - 1; h < i; h++ {
28+
f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
29+
}
30+
}
31+
}
32+
}
33+
return f[n][k]
34+
}
35+
36+
func min(a, b int) int {
37+
if a < b {
38+
return a
39+
}
40+
return b
41+
}

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