[leetcode] Largest Rectangle in Histogram

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.

https://oj.leetcode.com/problems/largest-rectangle-in-histogram/

 

思路1：穷举左右边界，O(n^2)，太慢。

思路2：解法见参考2。复杂度 O(n)。

　　该算法正确的原因：详见参考1，我的理解大概是这样：我们要做的是，对于任意的一个bar 'x'，我们需要计算以x为高度能形成的最大矩形。为了计算以x为高度的最大矩形，我们需要找到左边第一个比x矮的位置和右边一的哥比x矮的位置来计算这个矩形的宽度。对于这个用stack的巧妙解法，当遇到递减的bar时，我们pop并计算以这个bar为高度的最大矩形，遇到的递减的bar就是右边界，而左边界就是栈顶的元素。

 

public class Solution {

    public int largestRectangleArea(int[] height) {
        Stack<Integer> stack = new Stack<Integer>();
        int maxArea = 0;
        for (int i = 0; i < height.length;) {
            if (stack.isEmpty() || height[i] >= height[stack.peek()]) {
                stack.push(i++);
            } else {
                int start = stack.pop();
                int width = stack.isEmpty() ? i : (i - stack.peek() - 1);
                maxArea = Math.max(maxArea, height[start] * width);
            }

        }

        while (!stack.isEmpty()) {
            int start = stack.pop();
            int width = stack.isEmpty() ? height.length : (height.length - stack.peek() - 1);
            maxArea = Math.max(maxArea, height[start] * width);
        }

        return maxArea;
    }

    public static void main(String[] args) {
        System.out.println(new Solution().largestRectangleArea(new int[] { 2, 1, 5, 6, 2, 3 }));
    }
}

 

第二遍记录：算法不变。

　　注意stack里存储的是索引，所以比较的时候要height[idx]换算成高度。

　　注意当前bar比stack顶矮的时候，一直pop指导栈空或者比当前bar矮，i值不增加。

public class Solution {
    public int largestRectangleArea(int[] height) {
        Stack<Integer> stack = new Stack<Integer>();
        int maxArea = 0;
        
        for(int i=0;i<height.length;){
            if(stack.isEmpty()||height[i]>=height[stack.peek()]){
                stack.push(i++);
            }else{
                int curIdx = stack.pop();
                int width = stack.isEmpty()?(i):(i-stack.peek()-1);
                maxArea = Math.max(maxArea, height[curIdx] * width);
            }
        }
        
        while(!stack.isEmpty()){
            int curIdx = stack.pop();
            int width = stack.isEmpty() ? height.length : (height.length - stack.peek() - 1);
            maxArea = Math.max(maxArea, height[curIdx] * width);
            
        }
        return maxArea;
    }
}

 

2022， 同样的算法，可以开始push -1 到stack底下。

class Solution {
    public int largestRectangleArea(int[] heights) {
        Stack<Integer> stack = new Stack<>();
        stack.push(-1);
        
        int n = heights.length;
        int maxArea = 0;
        
        for(int i=0;i<n;i++){
            while(stack.peek()!=-1 && heights[stack.peek()]>=heights[i]){
                int curHeight = heights[stack.pop()];
                int curWidth = i-stack.peek()-1;
                maxArea = Math.max(maxArea,curHeight*curWidth);
            }
            
            stack.push(i);
        }
        while(stack.peek()!=-1){
            int curHeight = heights[stack.pop()];
            int curWidth = n - stack.peek() - 1;
            maxArea = Math.max(maxArea, curHeight*curWidth);
        }
        return maxArea;
    }
}

　　

 

 

参考：

http://www.geeksforgeeks.org/largest-rectangle-under-histogram/

http://www.cnblogs.com/lichen782/p/leetcode_Largest_Rectangle_in_Histogram.html#2973562

http://blog.csdn.net/linhuanmars/article/details/20524507