LeetCode 84. Largest Rectangle in Histogram

原题链接在这里：https://leetcode.com/problems/largest-rectangle-in-histogram/description/

题目：

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.

题解： 

用stack保存数组的index, 确保stack中index对应的高度是递增的.

当出现较小高度时需要不停的pop出比它大的index, 更新最大面积. 

如何更新呢. 高度是heights[stk.pop()], 宽度是stk pop后的顶部index, stk.peek()到i之间的距离-1. 

Because if there is higher bar before current index, it is no longer in the stack, the the width needs to cover that.

e.g. [1, 4, 3, 2]. When it comes to 2, the stack pop index 2, then stack top index becomes 0. height 3, its width should be 2. index 3 - top 0 - 1.

最后在从后到前更新次最大面积.

Time Complexity: O(n). n = heights.length. 

Space: O(n).

AC Java:

class Solution {
    public int largestRectangleArea(int[] heights) {
        if(heights == null || heights.length == 0){
            return 0;
        }
        
        int res = 0;
        Stack<Integer> stk = new Stack<Integer>();
        stk.push(-1);
        
        for(int i = 0; i<heights.length; i++){
            while(stk.peek()!=-1 && heights[stk.peek()]>=heights[i]){
                int h = heights[stk.pop()];
                res = Math.max(res, h*(i-stk.peek()-1));        
            }
            
            stk.push(i);
        }
        
        while(stk.peek()!=-1){
            res = Math.max(res, heights[stk.pop()]*(heights.length-stk.peek()-1));
        }
                           
        return res;        
    }
}

 跟上Maximal Rectangle.