classSolution {public:voidsetZeroes(vector<vector<int> > &matrix) {intlen1 =matrix.size();if(len1 ==0)return;intlen2 = matrix[0].size();if(len2 ==0)return;introw = -1, col = -1;for(inti =0; i < len1; i++)for(intj =0; j < len2; j++) {if(matrix[i][j] ==0)...
1voidsetZeroes(vector<vector<int>>&matrix) {2intn = matrix.size(), m = matrix[0].size();3int*row = (int*)malloc((n /31+2) *sizeof(int));4int*col = (int*)malloc((m /31+2) *sizeof(int));57for(inti =0; i < n /31+2; i ++) row[i] =0;8for(inti =0; i < ...
Can you solve this real interview question? Set Matrix Zeroes - Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's. You must do it in place [https://en.wikipedia.org/wiki/In-place_algorithm]. Example 1: [
Can you solve this real interview question? Set Matrix Zeroes - Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's. You must do it in place [https://en.wikipedia.org/wiki/In-place_algorithm]. Example 1: [
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. 对matrixi==0的部分,对i行、j列全部原地置为0. 2. 思路 找到第一个0的位置,将这个位置的行和列作为标记数组。 3. 代码 耗时:53ms ...
public void setZeroes(int[][] matrix) { int m = matrix.length, n = matrix[0].length; int row0 = 1, column0 = 1; // first row for (int i = 0; i < n; i++) { if (0 == matrix[0][i]) { row0 = 0; } }
Set Matrix Zeroes Given amxnmatrix, 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. ...
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. 提示: 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. ...
【Leetcode】Set Matrix Zeroes 给定一个m x n的矩阵,如果某个元素为0,则把该元素所在行和列全部置0。 Given amxnmatrix, if an element is 0, set its entire row and column to 0. Do it in place. 分析:最直接想到的就是遍历该矩阵,每遇到0就把它所在的行和列全部置0,但这是错的,因为这样会...
这道题中说的空间复杂度为O(mn)的解法自不用多说,直接新建一个和matrix等大小的矩阵,然后一行一行的扫,只要有0,就将新建的矩阵的对应行全赋0,行扫完再扫列,然后把更新完的矩阵赋给matrix即可,这个算法的空间复杂度太高。将其优化到O(m+n)的方法是,用一个长度为m的一维数组记录各行中是否有0,用一个...