leetcode85.最大矩形

leetcode85.最大矩形

题目

给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。

用例

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：6
解释：最大矩形如上图所示。

输入：matrix = []
输出：0

输入：matrix = [["0"]]
输出：0

输入：matrix = [["1"]]
输出：1

输入：matrix = [["0","0"]]
输出：0

求解

/**
 * @param {character[][]} matrix
 * @return {number}
 */
var maximalRectangle = function(matrix) {
    let max_Area = 0
    for(let i=0;i<matrix.length;i++){
        let heights = []
        for(j=0;j<matrix[0].length;j++){
            let height = 0
            let p=i;
            while(p>=0){
                if(matrix[p][j]==0){
                    break
                }
                height++
                p--
            }
            heights.push(height)
        }
        let tmp = largestArea(heights)
        tmp>max_Area?max_Area=tmp:max_Area=max_Area 
    }

    return max_Area


    function largestArea(heights){
        let height_stack = []
        let index_stack = []
        let maxArea = 0
        //哨兵
        heights[heights.length]=0
        height_stack.push(heights[0])
        index_stack.push(0)
        let final_index = 0
        let flag = false
        let i =1
        while(i<heights.length){
            //如果第i个柱子高度要严格小于栈顶元素，那么以栈顶元素为高的矩阵可以计算出其高度
            if(heights[i]<height_stack[height_stack.length-1]){
                flag=true
                //出栈
                let height = height_stack[height_stack.length-1]
                height_stack.pop()
                let index = index_stack[index_stack.length-1]
                final_index =index
                index_stack.pop()
                //计算宽度
                let weight = i-index
                weight*height>maxArea?maxArea=weight*height:maxArea=maxArea
            }else if(heights[i]==height_stack[height_stack.length-1]){
                flag=false
                //相等的情况直接无视
                i++
            }else{
                //入栈
                //如果是出栈以后的入栈
                if(flag==true){
                    height_stack.push(heights[i])
                    index_stack.push(final_index)
                    flag=false
                }else{
                    height_stack.push(heights[i])
                    index_stack.push(i)
                }
                i++            
            }
        }
        return maxArea
    }
};