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，刚開始想的双指针。前后扫描，可是每次求最小的高度的时候都须要遍历一次，效率上不是非常高。为o(n2)。

代码例如以下：

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;

        		//又一次得到最小值

        		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){//假设相等，则结束

            	break;

            }

            if(height[i] < height[j]){//左指针添加，直到比当前大

            	int k = i;

            	while(k <= j && height[k] <= height[i]){

            		if(height[k] == minHeight){//推断最小值是否改变

            			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){//推断最小值是否改变

            			isMinChange = true;

            		}

            		k--;

            	}

            	j = k;

            }

        }

        return max;

    }

}

解法上过不了OJ,所以仅仅能在网上參看资料。最后找到资料例如以下。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;

    }

}