Leetcode 84. Largest Rectangle in Histogram

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3]
.

The largest rectangle is shown in the shaded area, which has area = 10
unit.

For example,Given heights = [2,1,5,6,2,3],return 10.

题意：有一个用数组表示的树状图，找出这个树状图中最大的矩形。

思路：自己想到的就是比较暴力的解法，尝试每个可能的矩形面积，找出最大的。这里面有一个优化可以做到n方时间复杂度，就是尝试每个起点时，用一个变量记录到当前终点位置，当前最小的高度是多少。

public int largestRectangleArea(int[] heights) {
    int maxarea = 0;
    for (int i = 0; i < heights.length; i++) {
        int minheight = Integer.MAX_VALUE;
        for (int j = i; j < heights.length; j++) {
            minheight = Math.min(minheight, heights[j]);
            maxarea = Math.max(maxarea, minheight * (j - i + 1));
        }
    }
    return maxarea;
}

想不出来线性的解法，看了答案，答案的思路是用栈维护一个递增序列，一旦碰到了一个小于栈顶的新元素，就知道了以当前栈顶元素高度为终点的最大矩形是多少了。

public int largestRectangleArea(int[] heights) {
    Stack < Integer > stack = new Stack < > ();
    stack.push(-1);
    int maxarea = 0;
    for (int i = 0; i < heights.length; ++i) {
        while (stack.peek() != -1 && heights[stack.peek()] >= heights[i])
            maxarea = Math.max(maxarea, heights[stack.pop()] * (i - stack.peek() - 1));
        stack.push(i);
    }
    while (stack.peek() != -1) {
        maxarea = Math.max(maxarea, heights[stack.pop()] * (heights.length - stack.peek() -1));
    }
    return maxarea;
}