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Commit 52cbdd9

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dp: max area
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/*
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* @lc app=leetcode.cn id=84 lang=cpp
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*
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* [84] 柱状图中最大的矩形
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*
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* https://leetcode.cn/problems/largest-rectangle-in-histogram/description/
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*
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* algorithms
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* Hard (44.98%)
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* Likes: 2446
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* Dislikes: 0
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* Total Accepted: 342.6K
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* Total Submissions: 761.7K
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* Testcase Example: '[2,1,5,6,2,3]'
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*
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* 给定 n 个非负整数,用来表示柱状图中各个柱子的高度。每个柱子彼此相邻,且宽度为
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* 1 。
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*
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* 求在该柱状图中,能够勾勒出来的矩形的最大面积。
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*
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*
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*
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* 示例 1:
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*
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*
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*
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*
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* 输入:heights = [2,1,5,6,2,3]
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* 输出:10
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* 解释:最大的矩形为图中红色区域,面积为 10
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*
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*
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* 示例 2:
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*
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*
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*
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*
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* 输入: heights = [2,4]
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* 输出: 4
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*
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*
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*
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* 提示:
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*
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*
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* 1
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* 0
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*
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*
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*/
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#include <iostream>
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#include <stack>
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#include <vector>
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using namespace std;
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// @lc code=start
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class Solution {
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public:
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// 每个高度对应的最大面积由宽度决定,宽度则为左右第一个比该高度小所包起来的区间
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// 举例[8,2,4,5,3,2], 直接看245,到3时则可以确定5的最大面积是5,然后4是4*2
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// 2则还不能确定, 所以,发现递增的时候先记录起来,当发现比最新值小,则可以计算
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// 显然可以用栈来记录,也就是单调栈
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int largestRectangleArea(vector<int> &heights) {
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int n = heights.size();
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if (n < 2) {
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return n < 1 ? 0 : heights[0];
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}
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stack<int> stk; // 存位置
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stk.push(0);
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int ret = heights[0];
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for (int i = 1; i < heights.size(); i++) {
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while (!stk.empty() && heights[i] < heights[stk.top()]) {
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int h = heights[stk.top()];
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stk.pop();
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// 左侧是栈的下一个元素,右侧是i,栈空了说明左侧没有更小,也就左为0,宽度i-0,比如5/4/1/2的5和4
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int w = stk.empty() ? i : (i - stk.top() - 1);
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ret = max(ret, w * h);
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}
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stk.push(i);
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}
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while (!stk.empty()) {
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int h = heights[stk.top()];
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stk.pop();
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int w = stk.empty() ? n : (n - stk.top() - 1);
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ret = max(ret, w * h);
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}
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return ret;
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}
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};
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// @lc code=end
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int main(int argc, const char **argv) {
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Solution s;
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vector<int> arr({5, 4, 1, 2});
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auto ret = s.largestRectangleArea(arr);
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std::cout << "ret: " << ret;
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return 0;
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}

‎cpp/leetcode/85.最大矩形.cpp

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/*
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* @lc app=leetcode.cn id=85 lang=cpp
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*
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* [85] 最大矩形
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*
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* https://leetcode.cn/problems/maximal-rectangle/description/
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*
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* algorithms
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* Hard (54.65%)
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* Likes: 1528
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* Dislikes: 0
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* Total Accepted: 175.6K
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* Total Submissions: 321.4K
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* Testcase Example:
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* '[["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]'
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*
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* 给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵,找出只包含 1
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* 的最大矩形,并返回其面积。
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*
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*
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*
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* 示例 1:
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*
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*
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* 输入:matrix =
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* [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
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* 输出:6
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* 解释:最大矩形如上图所示。
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*
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*
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* 示例 2:
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*
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*
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* 输入:matrix = []
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* 输出:0
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*
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*
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* 示例 3:
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*
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*
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* 输入:matrix = [["0"]]
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* 输出:0
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*
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*
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* 示例 4:
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*
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*
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* 输入:matrix = [["1"]]
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* 输出:1
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*
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*
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* 示例 5:
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*
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*
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* 输入:matrix = [["0","0"]]
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* 输出:0
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*
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*
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*
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*
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* 提示:
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*
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*
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* rows == matrix.length
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* cols == matrix[0].length
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* 1 <= row, cols <= 200
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* matrix[i][j] 为 '0' 或 '1'
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*
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*
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*/
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#include <iostream>
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#include <stack>
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#include <vector>
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using namespace std;
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// @lc code=start
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class Solution {
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public:
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// 可以看成是84题的2维版本,0-n行的最大面积
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int maximalRectangle(vector<vector<char>> &matrix) {
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int m = matrix.size();
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if (m == 0) {
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return 0;
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}
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int n = matrix[0].size();
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vector<vector<int>> heights(m, vector<int>(n, 0));
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (matrix[i][j] == '1') {
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heights[i][j] = (i == 0 ? 0 : heights[i - 1][j]) + 1;
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}
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}
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}
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int ret = 0;
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// 每一层按84题算一次
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for (int i = 0; i < m; i++) {
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ret = max(ret, largestRectangleArea(heights[i]));
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}
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return ret;
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}
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private:
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// 84题的最大面积
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int largestRectangleArea(vector<int> &heights) {
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int n = heights.size();
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if (n < 2) {
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return n < 1 ? 0 : heights[0];
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}
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stack<int> stk; // 存位置
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stk.push(0);
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int ret = heights[0];
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for (int i = 1; i < heights.size(); i++) {
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while (!stk.empty() && heights[i] < heights[stk.top()]) {
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int h = heights[stk.top()];
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stk.pop();
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// 左侧是栈的下一个元素,右侧是i,栈空了说明左侧没有更小,也就左为0,宽度i-0,比如5/4/1/2的5和4
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int w = stk.empty() ? i : (i - stk.top() - 1);
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ret = max(ret, w * h);
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}
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stk.push(i);
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}
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while (!stk.empty()) {
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int h = heights[stk.top()];
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stk.pop();
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int w = stk.empty() ? n : (n - stk.top() - 1);
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ret = max(ret, w * h);
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}
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return ret;
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}
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};
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// @lc code=end

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