题目

法1：单调栈[原版]

O(N)+O(N)
必须掌握算法！！！

class Solution {public int largestRectangleArea(int[] heights) {int n = heights.length, res = 0;int[] leftMin = new int[n], rightMin = new int[n];Stack<Integer> stack = new Stack<>();for (int i = 0; i < n; ++i) {while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {stack.pop();}leftMin[i] = stack.isEmpty() ? -1 : stack.peek();stack.push(i);}stack.clear();for (int i = n - 1; i >= 0; --i) {while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {stack.pop();}rightMin[i] = stack.isEmpty() ? n : stack.peek();stack.push(i);}for (int i = 0; i < n; ++i) {res = Math.max(res, (rightMin[i] - leftMin[i] - 1) * heights[i]);}return res;}
}

法2：单调栈[优化版]

O(N)+O(N)
参考答案

class Solution {public int largestRectangleArea(int[] heights) {int n = heights.length, res = 0;int[] leftMin = new int[n], rightMin = new int[n];Arrays.fill(rightMin, n); // 一定注意这次需要初始化！！！Stack<Integer> stack = new Stack<>();for (int i = 0; i < n; ++i) {while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {rightMin[stack.peek()] = i;stack.pop();}leftMin[i] = stack.isEmpty() ? -1 : stack.peek();stack.push(i);}for (int i = 0; i < n; ++i) {res = Math.max(res, (rightMin[i] - leftMin[i] - 1) * heights[i]);}return res;}
}