Leetcode 84. Largest Rectangle in Histogram 最大矩形 解题报告

1 解题思想

这道题是说给了一个直方图，然后需要在里面找出一个最大的矩形。

这道题的解题方式，就是要使用一个栈，保持一个持续上升的队列

具体的请参照这个：http://blog.csdn.net/doc_sgl/article/details/11805519

我是参照这个的

2 原题

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.

3 AC解

import java.util.*;

public class Solution {
    /**
     * http://blog.csdn.net/doc_sgl/article/details/11805519
     * 
     * 这道题关键是要连续上升的形状，还有注意len的计算位置
     * */
    public int largestRectangleArea(int[] heights) {
        Stack stack = new Stack();
        int max_area=0,currentHeight,lastIndex,len;
        for(int i=0;i//保持升序
            if(stack.isEmpty() || heights[stack.peek()]<=currentHeight){
                stack.push(i);
                continue;
            }
            //到了这里，就是不满足升序，需要弹出，注意当前len的判断方式
            while(stack.isEmpty()==false && heights[stack.peek()]>currentHeight){
                lastIndex=stack.pop();
                len=stack.isEmpty()?i:(i-stack.peek()-1);
                max_area=Math.max(max_area,len*heights[lastIndex]);
            }
            stack.push(i);
        }
        //最后需要多处理一次
        while(stack.isEmpty()==false){
            lastIndex=stack.pop();
            len=stack.isEmpty()?heights.length:(heights.length-stack.peek()-1);
            max_area=Math.max(max_area,len*heights[lastIndex]);
        }
        return max_area;

    }

}