题目：
给定一个仅包含 0 和 1 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。
示例:
输入:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
输出: 6

解法一 暴力法优化
思路：
1、将数组转化为存放当前连续宽度；
2、基于84暴力法优化，只是将一维数据改为二维；

图片来源力扣官网

时间复杂度 O(n^2m)
空间复杂度 O(1)

var maximalRectangle = function (matrix){
    if(!matrix.length) return 0;
    let max = 0;
    for(let i = 0, len = matrix.length; i < len; i++){
        for(let j = 0, jlen = matrix[0].length; j < jlen; j++){
            if(j === 0) 
                matrix[i][j] = parseInt(matrix[i][j]);
            else
                matrix[i][j] = matrix[i][j] === '1' ? matrix[i][j - 1] + 1 : 0;
            let min = matrix[i][j];
            for(let k = i; k >= 0; k--){
                min = Math.min(min, matrix[k][j])
                max = Math.max(max, (i - k + 1) * min);
            }
        }
        
    }
    return max
}

解法二 栈
思路：
1、基于84栈优化，只是将一维数据改为二维，并将矩阵转置；

时间复杂度 O(nm)

var maximalRectangle = function (matrix){
    if(!matrix.length) return 0;
    let max = 0;
    let sumMatrix = getSumMatrix(matrix);
    let triMatrix = getTriMatrix(sumMatrix);
    for(let i = 0, len = triMatrix.length; i < len; i++){
        max = Math.max(max, largestRectangleArea(triMatrix[i]))
    }
    return max
}
// 栈
var largestRectangleArea = function(heights) {
    let i = 0;
    let max = 0;
    let stack = [];
    heights.push(0);
    heights.unshift(0);
    while(i < heights.length){
        while(stack.length && heights[i] < heights[stack[stack.length - 1]]){
            max = Math.max(max, heights[stack.pop()] * (i - stack[stack.length - 1] - 1))
        }
        stack.push(i++);
    }
    return max;
};
var getSumMatrix = function(matrix) {
    return matrix.map(item => {
        let num = 0;
        let arr = [];
        item.forEach(i => {
            if(i === '1'){
                num += 1;
            }else{
                num = 0;
            }
            arr.push(num);
        })
        return arr;
    });
}
var getTriMatrix = function(matrix) {
    let result = [],
        len = matrix.length,
        childLen = matrix[0].length;
    for(let i = 0; i < childLen; i++){
        let arr = [];
        for(let j = 0; j < len; j++){
            arr.push(matrix[j][i])
        }
        result.push(arr);
    }
    return result;
}