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

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;
    }
}