[LeetCode][JavaScript]Set Matrix Zeroes

Set Matrix Zeroes

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

click to show follow up.

Follow up:

Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

https://leetcode.com/problems/set-matrix-zeroes/

---

数组中如果出现0，把这个数行和列上的数都置成0.

要求常数级的空间复杂度。

一开始想用二进制解决，开2个变量代表行和列。

把行号和列号做2的左移运算，然后用或操作记录，用与判断重复。

大数据越界，挂在倒数3个case上，数组长度有50。

第二种做法，把数组中的0所在的行和列都换成null，最后换回0就好了。

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes = function(matrix) {
    if(matrix.length > 0){
        var rowSize = matrix.length, colSize = matrix[0].length;
        var i, j, m, n;
        for(i = 0; i < rowSize; i++){
            for(j = 0; j < colSize; j++){
                if(matrix[i][j] === 0){
                    for(m = 0; m < rowSize; m++){
                        if(matrix[m][j] !== 0){
                            matrix[m][j] = null;
                        }
                    }
                    for(n = 0; n < colSize; n++){
                        if(matrix[i][n] !== 0){
                            matrix[i][n] = null;
                        }
                    }
                }                
            }
        }

        for(i = 0; i < rowSize; i++){
            for(j = 0; j < colSize; j++){
                if(matrix[i][j] === null){
                    matrix[i][j] = 0;
                }                
            }
        }
    }
};

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes_outOfBundary = function(matrix) {
    var row = 0, col = 0;
    if(matrix.length > 0){
        var rowSize = matrix.length, colSize = matrix[0].length;
        var i, j, curr, rowBinary, colBinary;
        for(i = 0; i < rowSize; i++){
            for(j = 0; j < colSize; j++){
                if(matrix[i][j] === 0){
                    rowBinary = 2 << i;
                    colBinary = 2 << j;                    
                    if((row & rowBinary) === 0){
                        row = row | rowBinary;
                    }
                    if((col & colBinary) === 0){
                        col = col | colBinary;
                    }
                }                
            }
        }

        for(i = 0; i < rowSize; i++){
            rowBinary = 2 << i;
            if((row & rowBinary) !== 0){
                for(j = 0; j < colSize; j++){
                    matrix[i][j] = 0;
                }
            }
        }

        for(j = 0; j < colSize; j++){
            colBinary = 2 << j;
            if((col & colBinary) !== 0){
                for(i = 0; i < rowSize; i++){
                    matrix[i][j] = 0;
                }
            }
        }
    }
};