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feat: add solutions to lcp problem: No.12
LCP 12.小张刷题计划
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‎lcp/LCP 12. 小张刷题计划/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 二分查找**
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我们可以将题意转换为，将题目最多分成 $m$ 组，每一组去掉最大值后不超过 $T$ ，求最小的满足条件的 $T$。
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我们定义二分查找的左边界 $left=0$，右边界 $right=\sum_{i=0}^{n-1}time[i]$，二分查找的目标值为 $T$。
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我们定义函数 $check(T)$，表示是否存在一种分组方案，使得每一组去掉最大值后不超过 $T$，并且分组数不超过 $m$。
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我们可以用贪心的方法来判断是否存在这样的分组方案。我们从左到右遍历题目，将题目耗时加入当前总耗时 $s$，并更新当前分组的最大值 $mx$。如果当前总耗时 $s$ 减去当前分组的最大值 $mx$ 大于 $T$，则将当前题目作为新的分组的第一题，更新 $s$ 和 $mx$。继续遍历题目，直到遍历完所有题目。如果分组数不超过 $m$，则说明存在这样的分组方案，返回 $true$，否则返回 $false$。
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时间复杂度 $O(n \times \log S)$，空间复杂度 $O(1)$。其中 $n$ 和 $S$ 分别为题目数量和题目总耗时。
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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 minTime(self, time: List[int], m: int) -> int:
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def check(t):
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s = mx = 0
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d = 1
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for x in time:
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s += x
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mx = max(mx, x)
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if s - mx > t:
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d += 1
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s = mx = x
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return d <= m
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return bisect_left(range(sum(time)), True, key=check)
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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 minTime(int[] time, int m) {
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int left = 0, right = 0;
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for (int x : time) {
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right += x;
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}
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while (left < right) {
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int mid = (left + right) >> 1;
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if (check(mid, time, m)) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private boolean check(int t, int[] time, int m) {
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int s = 0, mx = 0;
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int d = 1;
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for (int x : time) {
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s += x;
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mx = Math.max(mx, x);
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if (s - mx > t) {
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s = x;
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mx = x;
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++d;
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}
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}
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return d <= m;
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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 minTime(vector<int>& time, int m) {
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int left = 0, right = 0;
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for (int x : time) {
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right += x;
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}
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auto check = [&](int t) -> bool {
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int s = 0, mx = 0;
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int d = 1;
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for (int x : time) {
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s += x;
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mx = max(mx, x);
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if (s - mx > t) {
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s = x;
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mx = x;
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++d;
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}
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}
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return d <= m;
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};
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while (left < right) {
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int mid = (left + right) >> 1;
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if (check(mid)) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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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 minTime(time []int, m int) int {
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right := 0
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for _, x := range time {
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right += x
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}
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return sort.Search(right, func(t int) bool {
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s, mx := 0, 0
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d := 1
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for _, x := range time {
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s += x
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mx = max(mx, x)
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if s-mx > t {
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s, mx = x, x
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d++
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}
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}
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return d <= m
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})
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}
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func max(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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### **...**

‎lcp/LCP 12. 小张刷题计划/Solution.cpp

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class Solution {
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public:
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int minTime(vector<int>& time, int m) {
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int left = 0, right = 0;
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for (int x : time) {
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right += x;
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}
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auto check = [&](int t) -> bool {
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int s = 0, mx = 0;
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int d = 1;
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for (int x : time) {
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s += x;
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mx = max(mx, x);
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if (s - mx > t) {
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s = x;
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mx = x;
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++d;
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}
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}
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return d <= m;
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};
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while (left < right) {
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int mid = (left + right) >> 1;
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if (check(mid)) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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};

‎lcp/LCP 12. 小张刷题计划/Solution.go

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func minTime(time []int, m int) int {
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right := 0
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for _, x := range time {
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right += x
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}
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return sort.Search(right, func(t int) bool {
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s, mx := 0, 0
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d := 1
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for _, x := range time {
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s += x
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mx = max(mx, x)
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if s-mx > t {
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s, mx = x, x
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d++
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}
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}
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return d <= m
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})
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}
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func max(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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}

‎lcp/LCP 12. 小张刷题计划/Solution.java

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class Solution {
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public int minTime(int[] time, int m) {
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int left = 0, right = 0;
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for (int x : time) {
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right += x;
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}
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while (left < right) {
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int mid = (left + right) >> 1;
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if (check(mid, time, m)) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private boolean check(int t, int[] time, int m) {
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int s = 0, mx = 0;
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int d = 1;
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for (int x : time) {
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s += x;
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mx = Math.max(mx, x);
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if (s - mx > t) {
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s = x;
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mx = x;
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++d;
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}
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}
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return d <= m;
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}
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}

‎lcp/LCP 12. 小张刷题计划/Solution.py

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class Solution:
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def minTime(self, time: List[int], m: int) -> int:
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def check(t):
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s = mx = 0
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d = 1
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for x in time:
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s += x
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mx = max(mx, x)
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if s - mx > t:
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d += 1
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s = mx = x
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return d <= m
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return bisect_left(range(sum(time)), True, key=check)

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