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✨update: Modify 363、611 #486

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2 changes: 1 addition & 1 deletion Index/双指针.md
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Expand Up @@ -36,7 +36,7 @@
| [1004. 最大连续1的个数 III](https://leetcode-cn.com/problems/max-consecutive-ones-iii/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/max-consecutive-ones-iii/solution/san-chong-jie-fa-cong-dong-tai-gui-hua-d-gxks/) | 中等 | 🤩🤩🤩 |
| [1052. 爱生气的书店老板](https://leetcode-cn.com/problems/grumpy-bookstore-owner/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/grumpy-bookstore-owner/solution/hua-dong-chuang-kou-luo-ti-by-ac_oier-nunu/) | 中等 | 🤩🤩🤩 |
| [1221. 分割平衡字符串](https://leetcode-cn.com/problems/split-a-string-in-balanced-strings/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/split-a-string-in-balanced-strings/solution/gong-shui-san-xie-noxiang-xin-ke-xue-xi-wumnk/) | 简单 | 🤩🤩🤩🤩 |
| [1332. 删除回文子序列](https://leetcode-cn.com/problems/remove-palindromic-subsequences/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/remove-palindromic-subsequences/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-0zwn/) | 中等 | 🤩🤩🤩🤩 |
| [1332. 删除回文子序列](https://leetcode-cn.com/problems/remove-palindromic-subsequences/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/remove-palindromic-subsequences/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-0zwn/) | 简单 | 🤩🤩🤩🤩 |
| [1446. 连续字符](https://leetcode-cn.com/problems/consecutive-characters/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/consecutive-characters/solution/gong-shui-san-xie-jian-dan-shuang-zhi-zh-xtv6/) | 简单 | 🤩🤩🤩🤩🤩 |
| [1610. 可见点的最大数目](https://leetcode-cn.com/problems/maximum-number-of-visible-points/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/maximum-number-of-visible-points/solution/gong-shui-san-xie-qiu-ji-jiao-ji-he-ti-b-0bid/) | 困难 | 🤩🤩🤩🤩 |
| [1743. 从相邻元素对还原数组](https://leetcode-cn.com/problems/restore-the-array-from-adjacent-pairs/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/restore-the-array-from-adjacent-pairs/solution/gong-shui-san-xie-yi-ti-shuang-jie-dan-x-elpx/) | 中等 | 🤩🤩🤩🤩 |
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163 changes: 144 additions & 19 deletions LeetCode/361-370/363. 矩形区域不超过 K 的最大数值和（困难）.md
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Expand Up @@ -6,13 +6,13 @@ Tag : 「二分」、「前缀和」



给你一个 m x n 的矩阵 matrix 和一个整数 k ，找出并返回矩阵内部矩形区域的不超过 k 的最大数值和。
给你一个 `m x n` 的矩阵 `matrix` 和一个整数 $k$ ，找出并返回矩阵内部矩形区域的不超过 $k$ 的最大数值和。

题目数据保证总会存在一个数值和不超过 k 的矩形区域。


题目数据保证总会存在一个数值和不超过 $k$ 的矩形区域。

示例 1：
![](https://assets.leetcode.com/uploads/2021/03/18/sum-grid.jpg)

```
输入：matrix = [[1,0,1],[0,-2,3]], k = 2

Expand All @@ -28,11 +28,11 @@ Tag : 「二分」、「前缀和」
```

提示：
* m == matrix.length
* n == matrix[i].length
* 1 <= m, n <= 100
* -100 <= matrix[i][j] <= 100
* -$10^5$ <= k <= $10^5$
* $m == matrix.length$
* $n == matrix[i].length$
* $1 <= m, n <= 100$
* $-100 <= matrix[i][j] <= 100$
* -$10^5 <= k <= 10^5$

---

Expand All @@ -46,8 +46,8 @@ Tag : 「二分」、「前缀和」

数据范围是 $10^2$，对应的计算量是 $10^8$，理论上会超时，但当我们枚举「矩形左上角」$(i,j)$ 的时候，我们只需要搜索位于 $(i,j)$ 的右下方的点 $(p,q)$ 作为「矩形右下角」，所以其实我们是取不满 $m^2 * n^2$ 的，但仍然具有超时风险（2021/04/20 Java 测试可通过，C++ 使用 `vector` 会 TLE）。

代码：
```Java []
Java 代码：
```Java
class Solution {
public int maxSumSubmatrix(int[][] mat, int k) {
int m = mat.length, n = mat[0].length;
Expand All @@ -74,6 +74,35 @@ class Solution {
}
}
```
C++ 代码：
```C++
int sum[110][110];
class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& mat, int k) {
int m = mat.size(), n = mat[0].size();
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + mat[i - 1][j - 1];
}
}
int ans = INT_MIN;
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
for(int p = i; p <= m; p++){
for(int q = j; q <= n; q++){
int cur = sum[p][q] - sum[i - 1][q] - sum[p][j - 1] + sum[i - 1][j - 1];
if(cur <= k){
ans = max(ans,cur);
}
}
}
}
}
return ans;
}
};
```
* 时间复杂度：预处理前缀和数组复杂度为 $O(m * n)$，查找答案的复杂度为 $O(m^2 * n^2)$。整体复杂度为 $O(m^2 * n^2)$。
* 空间复杂度：$O(m * n)$

Expand Down Expand Up @@ -149,8 +178,8 @@ $$

至此，我们通过预处理前缀和 + 容斥原理彻底将题目转化为「一维问题」进行来求解。

代码：
```Java []
Java 代码：
```Java
class Solution {
public int maxSumSubmatrix(int[][] mat, int k) {
int m = mat.length, n = mat[0].length;
Expand Down Expand Up @@ -190,6 +219,38 @@ class Solution {
}
}
```
C++ 代码：
```C++
class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& mat, int k) {
int m = mat.size(), n = mat[0].size();
vector<vector<int>> sum(m + 1,vector<int>(n + 1,0));
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + mat[i - 1][j - 1];
}
}
int ans = INT_MIN;
for(int top = 1; top <= m; top++){
for(int bot = top; bot <= m; bot++){
set<int> st;
st.insert(0);
for(int r = 1; r <= n; r++){
int right = sum[bot][r] - sum[top - 1][r];
auto left = st.lower_bound(right - k);
if(left != st.end()){
int cur = right - *left;
ans = max(ans,cur);
}
st.insert(right);
}
}
}
return ans;
}
};
```
* 时间复杂度：枚举上下边界复杂度为 $O(m^2)$；枚举右边界为 $O(n)$，使用 `TreeSet`（基于红黑树）存储和查找左边界复杂度为 $O(\log{n})$。整体复杂度为 $O(m^2 * n\log{n})$
* 空间复杂度：$O(m * n)$

Expand All @@ -201,8 +262,8 @@ class Solution {

事实上，我们需要将「二分过程」应用到数值较大的行或者列之中，这样才能最大化我们查找的效率（同时也回答了本题的进阶部分）。

代码：
```Java []
Java 代码：
```Java
class Solution {
public int maxSumSubmatrix(int[][] mat, int k) {
int m = mat.length, n = mat[0].length;
Expand Down Expand Up @@ -234,6 +295,39 @@ class Solution {
}
}
```
C++ 代码：
```C++
class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& mat, int k) {
int m = mat.size(), n = mat[0].size();
vector<vector<int>> sum(m + 1,vector<int>(n + 1,0));
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + mat[i - 1][j - 1];
}
}
bool isRight = n > m;
int ans = INT_MIN;
for(int i = 1; i <= (isRight ? m : n); i++){
for(int j = i; j <= (isRight ? m : n); j++){
set<int> st;
st.insert(0);
for(int fixed = 1; fixed <= (isRight ? n : m); fixed++){
int a = isRight ? sum[j][fixed] - sum[i - 1][fixed] : sum[fixed][j] - sum[fixed][i - 1];
auto b = st.lower_bound(a - k);
if(b != st.end()){
int cur = a - *b;
ans = max(ans,cur);
}
st.insert(a);
}
}
}
return ans;
}
};
```
* 时间复杂度：预处理「每行」或「每列」的前缀和，复杂度为 $O(m * n)$；枚举子矩阵的「上下行」或「左右行」，复杂度为 $O(\min(m, n)^2)$；结合二维前缀和套用一维最大连续子数组解决方案，复杂度为$\max(m, n)\log{\max(m, n)}$。整体复杂度为 $O(\min(m, n)^2 * \max(m, n)\log{\max(m, n)})$
* 空间复杂度：$O(m * n)$

Expand All @@ -245,8 +339,8 @@ class Solution {

因此我们可以将计算前缀和的逻辑下放到搜索子矩阵的循环里去做，从而将 $O(m * n)$ 的空间复杂度下降到 $O(\max(m,n))$。

代码：
```Java []
Java 代码：
```Java
class Solution {
public int maxSumSubmatrix(int[][] mat, int k) {
int m = mat.length, n = mat[0].length;
Expand Down Expand Up @@ -275,18 +369,49 @@ class Solution {
}
}
```
C++ 代码：
```C++
class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& mat, int k) {
int m = mat.size(), n = mat[0].size();
bool isRight = n > m;
vector<int> sum((isRight ? n + 1 : m + 1), 0);
int ans = INT_MIN;
for(int i = 1; i <= (isRight ? m : n); i++){
fill(sum.begin(),sum.end(),0);
for(int j = i; j <= (isRight ? m : n); j++){
set<int> st;
st.insert(0);
int a = 0;
for(int fixed = 1; fixed <= (isRight ? n : m); fixed++){
sum[fixed] += isRight ? mat[j - 1][fixed - 1] : mat[fixed - 1][j - 1];
a += sum[fixed];
auto b = st.lower_bound(a - k);
if(b != st.end()){
int cur = a - *b;
ans = max(ans,cur);
}
st.insert(a);
}
}
}
return ans;
}
};
```
* 时间复杂度：预处理「每行」或「每列」的前缀和，复杂度为 $O(m * n)$；枚举子矩阵的「上下行」或「左右行」，复杂度为 $O(\min(m, n)^2)$；结合二维前缀和套用一维最大连续子数组解决方案，复杂度为$\max(m, n)\log{max(m, n)}$。整体复杂度为 $O(\min(m, n)^2 * \max(m, n)\log{\max(m, n)})$
* 空间复杂度：$O(\max(m, n))$

---

### 最后

这是我们「刷穿 LeetCode」系列文章的第 `No.363` 篇，系列开始于 2021/01/01，截止于起始日 LeetCode 上共有 1916 道题目，部分是有锁题，我们将先将所有不带锁的题目刷完。
这是我们「刷穿 LeetCode」系列文章的第 `No.363` 篇，系列开始于 2021/01/01，截止于起始日 LeetCode 上共有 1916 道题目，部分是有锁题，我们将先把所有不带锁的题目刷完。

在这个系列文章里面，除了讲解解题思路以外，还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。

为了方便各位同学能够电脑上进行调试和提交代码，我建立了相关的仓库：https://github.com/SharingSource/LogicStack-LeetCode。
为了方便各位同学能够电脑上进行调试和提交代码，我建立了相关的仓库：https://github.com/SharingSource/LogicStack-LeetCode 。

在仓库地址里，你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。

4 changes: 2 additions & 2 deletions LeetCode/611-620/611. 有效三角形的个数（中等）.md
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Expand Up @@ -21,8 +21,8 @@ Tag : 「排序」、「二分」、「双指针」
2,2,3
```
注意:
1. 数组长度不超过1000。
2. 数组里整数的范围为 [0, 1000]。
1. 数组长度不超过 $1000$。
2. 数组里整数的范围为 $[0, 1000]$。

---

Expand Down