leetCode 84.Largest Rectangle in Histogram （最大矩形直方图） 解题思路和方法

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 height = [2,1,5,6,2,3],
return 10.

思路：此题还是有一些难度的，刚开始想的双指针，前后扫描，但是每次求最小的高度的时候都需要遍历一次，效率上不是很高。为o(n2)。

代码如下：

public class Solution {    public int largestRectangleArea(int[] height) {            int max = 0;//最大值        int i = 0;//左指针        int j = height.length - 1;//右指针        boolean isMinChange = true;        //双指针扫描        while(i <= j){        int minHeight = Integer.MAX_VALUE;//范围内最小值        if(isMinChange){//最小值是否改变        isMinChange = false;        //重新得到最小值        for(int k = i ; k <= j;k++){                    if(height[k] < minHeight){                        minHeight = height[k];                    }                }        }        //面积            int area = (j - i + 1)*minHeight;            if(max < area){                max = area;            }            if(i == j){//如果相等，则结束            break;            }            if(height[i] < height[j]){//左指针增加，直到比当前大            int k = i;            while(k <= j && height[k] <= height[i]){            if(height[k] == minHeight){//判断最小值是否改变            isMinChange = true;            }            k++;            }                i = k;            }else{//右指针减小，直到比当前大            int k = j;            while( k >= i && height[k] <= height[j]){            if(height[k] == minHeight){//判断最小值是否改变            isMinChange = true;            }            k--;            }            j = k;            }        }         return max;    }}

解法上过不了OJ,所以只能在网上参看资料，最后找到资料如下，是用栈解决的。

public class Solution {    public int largestRectangleArea(int[] height) {    if (height == null || height.length == 0) return 0;            Stack<Integer> stHeight = new Stack<Integer>();        Stack<Integer> stIndex = new Stack<Integer>();        int max = 0;                for(int i = 0; i < height.length; i++){            if(stHeight.isEmpty() || height[i] > stHeight.peek()){                stHeight.push(height[i]);                stIndex.push(i);            }else if(height[i] < stHeight.peek()){            int lastIndex = 0;            while(!stHeight.isEmpty() && height[i] < stHeight.peek()){            lastIndex = stIndex.pop();            int area = stHeight.pop()*(i - lastIndex);            if(max < area){            max = area;            }            }            stHeight.push(height[i]);            stIndex.push(lastIndex);            }        }        while(!stHeight.isEmpty()){        int area = stHeight.pop()*(height.length - stIndex.pop());        max = max < area ? area:max;        }return max;    }}