【LeetCode】73. 矩阵置零

73. 矩阵置零

ts

/**
 Do not return anything, modify matrix in-place instead.
 */
function setZeroes(matrix: number[][]): void {
    const len1 = matrix.length
    const len2 = matrix[0].length
    let set1 = new Set([])
    let set2 = new Set([])
    for (let i1 = 0; i1 < len1; i1++) {
        for (let i2 = 0; i2 < len2; i2++) {
            if (matrix[i1][i2] === 0) {
                set1.add(i1)
                set2.add(i2)
            }
        }
    }
    for (let x1 of set1.values()) {
        for (let i2 = 0; i2 < len2; i2++) {
            matrix[x1][i2] = 0
        }
    }
    for (let x2 of set2.keys()) {
        for (let i1 = 0; i1 < len1; i1++) {
            matrix[i1][x2] = 0
        }
    }
};
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py

class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        len1 = len(matrix)
        len2 = len(matrix[0])
        arr1 = []
        arr2 = []
        for i1 in range(len1):
            for i2 in range(len2):
                if matrix[i1][i2] == 0:
                    arr1.append(i1)
                    arr2.append(i2)
        for x1 in arr1:
            for i2 in range(len2):
                matrix[x1][i2] = 0
        for x2 in arr2:
            for i1 in range(len1):
                matrix[i1][x2] = 0
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