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].

 

关键是转换视角，把问题看成求每个以heights[i]为高的矩形的面积，这样就能得到下面的思路。

public int largestRectangleArea(int[] heights) {
        int maxarea=0;
        Stack<Integer> stack=new Stack<>();
        for(int i=0;i<heights.length;i++){
            if(stack.empty()||heights[i]>heights[stack.peek()]){
                stack.push(i);
            }else{                
                while(!stack.empty()&&heights[stack.peek()]>heights[i]){//当前最高
                    maxarea=Math.max(heights[stack.pop()]*(stack.empty()?i:(i-stack.peek()-1)),maxarea);//中间隔了i-stack.peek()-1
                }
                stack.push(i);
            }   
        }   
        while(!stack.empty()){
            maxarea=Math.max(heights[stack.pop()]*(stack.empty()?heights.length:(heights.length-1-stack.peek())),maxarea);
        }
        return maxarea;
    }

 

 

leetcode讨论区看到一种更容易理解，简单的方法，类似回文子串方法https://leetcode.com/problems/largest-rectangle-in-histogram/discuss/28902/5ms-O(n)-Java-solution-explained-(beats-96)

定义两个数组，分别记录为面积>=当前面积的最左边下标和最右边下标，先遍历求出两个数组，每点扫描可以利用前一个点的扫描结果，使时间复杂度从O（n^2）降到了O(n)。最后跟遍历一遍根据上述两个数组记录的最左最右两边的下标，求出包括当前点的最大面积。

类似题目https://leetcode.com/problems/sum-of-subarray-minimums/