84. Largest Rectangle in Histogram

最新推荐文章于 2021-03-05 17:40:26 发布

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本文介绍了一种使用栈解决直方图中寻找最大矩形面积问题的方法。通过不断压栈与弹栈操作，计算每个柱状图所能形成的最大矩形面积。

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.

分析:通过栈来实现,栈中存储的是数组下标,当当前元素的值大于栈顶对应元素的值时,将当前下标压栈;当当前元素的值小于其值时,将其弹出,并依次计算弹出的值对应的矩阵的最大值.然后将当前元素下标压栈.

<span style="font-size:14px;">public class Solution {
    public int largestRectangleArea(int[] heights) {
        Stack<Integer> stack=new Stack<Integer>();
        int n=heights.length;
        int max=0,temp;
        for(int i=0;i<n;i++){
            if(stack.isEmpty()||heights[i]>=heights[stack.peek()]){
                stack.push(i);
            }else{
               
                while(!stack.isEmpty()&&heights[i]<heights[stack.peek()]){
                    temp=stack.pop();
                    if(!stack.isEmpty())
                        max=(i-stack.peek()-1)*heights[temp]>max?(i-stack.peek()-1)*heights[temp]:max;
                    else
                        max=i*heights[temp]>max?i*heights[temp]:max;
                }
                i--;
            }
        }
        if(!stack.isEmpty()){
            int index=stack.peek();
            while(!stack.isEmpty()){
                temp=stack.pop();
                if(!stack.isEmpty())
                        max=(index-stack.peek())*heights[temp]>max?(index-stack.peek())*heights[temp]:max;
                else
                        max=(index+1)*heights[temp]>max?(index+1)*heights[temp]:max;
            }
        }
        return max;
        
    }
}</span>