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Day39 矩阵置零

给定一个 m * n 的矩阵,如果一个元素为 0,则将其所在行和列的所有元素都设为 0。请使用原地算法

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

示例1:

示例2:

提示:

Java解法

package sj.shimmer.algorithm.m2;

import java.util.HashSet;
import java.util.Set;

import sj.shimmer.algorithm.Utils;

/**
 * Created by SJ on 2021/3/4.
 */

class D39 {
    public static void main(String[] args) {
        int[][] matrix = {{0,1,2,0}, {3,4,5,2}, {1,3,1,5}};
        setZeroes(matrix);
        Utils.logArrays(matrix);
    }
    public static void setZeroes(int[][] matrix) {
        if (matrix==null||matrix.length==0||matrix[0]==null||matrix[0].length==0) {
            return;
        }
        Set<Integer> zeroColumn = new HashSet<>();
        Set<Integer> zeroLine = new HashSet<>();
        int lineLength = matrix.length;
        int columnLength = matrix[0].length;
        for (int i = 0; i < lineLength; i++) {
            for (int j = 0; j < columnLength; j++) {
                if (matrix[i][j]==0) {
                    zeroLine.add(i);
                    zeroColumn.add(j);
                }
            }
        }
        for (int i = 0; i < lineLength; i++) {
            if (zeroLine.contains(i)) {
                matrix[i] = new int[columnLength];
                continue;
            }
            for (Integer integer : zeroColumn) {
                matrix[i][integer] = 0;
            }
        }
    }
}

官方解

https://leetcode-cn.com/problems/set-matrix-zeroes/solution/ju-zhen-zhi-ling-by-leetcode/

  1. 额外存储空间方法

    • 时间复杂度:O(M×N)
  • 空间复杂度:O(M+N)
  1. O(1)空间的暴力

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