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Commit 1ba26d0

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solve set_matrix_zeroes
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2 files changed

+163
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lines changed

‎algs/0073.set_matrix_zeros.go‎

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,96 @@
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package algs
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// SetZeroes
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//
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// Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's.
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//
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// You must do it in place.
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//
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// Example 1:
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//
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// Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
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// Output: [[1,0,1],[0,0,0],[1,0,1]]
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//
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// Example 2:
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//
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// Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
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// Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
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//
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// Constraints:
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//
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// m == matrix.length
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// n == matrix[0].length
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// 1 <= m, n <= 200
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// -231 <= matrix[i][j] <= 231 - 1
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//
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// Follow up:
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//
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// A straightforward solution using O(mn) space is probably a bad idea.
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// A simple improvement uses O(m + n) space, but still not the best solution.
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// Could you devise a constant space solution?
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func SetZeroes(matrix [][]int) {
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// // 直观解法
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// m := len(matrix)
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// n := len(matrix[0])
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// zeros := [2]map[int]bool {
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// 0: make(map[int]bool),
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// 1: make(map[int]bool),
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// }
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// for i := 0; i < m; i++ {
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// for j := 0; j < n; j++ {
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// if matrix[i][j] == 0 {
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// zeros[0][i] = true
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// zeros[1][j] = true
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// }
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// }
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// }
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// for i := 0; i < m; i++ {
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// for j := 0; j < n; j++ {
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// if zeros[0][i] || zeros[1][j] {
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// matrix[i][j] = 0
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// }
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// }
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// }
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// 空间复杂度为 O(1) 的解法
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// 使用 matrix 的初行初列代替上述 map 存储待置零行列标记
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//
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//
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// 使用初行初列标记那些行哪些列应该等于0
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var rowZero, columZero bool
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for i := 0; i < len(matrix); i++ {
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for j := 0; j < len(matrix[i]); j++ {
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if matrix[i][j] == 0 {
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matrix[i][0] = 0
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matrix[0][j] = 0
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if i == 0 {
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// 标记第一行中是否有 0
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rowZero = true
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}
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if j == 0 {
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// 标记第一列中是否有0
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columZero = true
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}
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}
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}
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}
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for i := 1; i < len(matrix); i++ {
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for j := 1; j < len(matrix[i]); j++ {
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if matrix[i][0] == 0 || matrix[0][j] == 0 {
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matrix[i][j] = 0
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}
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}
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}
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if rowZero {
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for i := 0; i < len(matrix[0]); i++ {
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matrix[0][i] = 0
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}
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}
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if columZero {
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for i := 0; i < len(matrix); i++ {
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matrix[i][0] = 0
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}
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}
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}

‎src/solution.rs‎

Lines changed: 67 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -874,7 +874,7 @@ impl Solution {
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result
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}
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/// [41. Frist Missing Positive]
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/// [41. Frist Missing Positive](https://leetcode.cn/problems/first-missing-positive)
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///
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/// Given an unsorted integer array nums. Return the smallest positive integer that is not present in nums.
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///
@@ -932,4 +932,70 @@ impl Solution {
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}
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nums.len() as i32 + 1
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}
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/// [73. Set Matrix Zeros](https://leetcode.cn/problems/set-matrix-zeroes/description)
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///
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/// Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's.
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///
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/// You must do it in place.
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///
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/// Example 1:
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///
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/// Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
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/// Output: [[1,0,1],[0,0,0],[1,0,1]]
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///
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/// Example 2:
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///
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/// Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
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/// Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
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///
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/// Constraints:
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///
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/// m == matrix.length
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/// n == matrix[0].length
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/// 1 <= m, n <= 200
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/// -231 <= matrix[i][j] <= 231 - 1
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///
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/// Follow up:
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///
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/// A straightforward solution using O(mn) space is probably a bad idea.
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/// A simple improvement uses O(m + n) space, but still not the best solution.
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/// Could you devise a constant space solution?
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pub fn set_zeros(matrix: &mut Vec<Vec<i32>>) {
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// 标记第一行/第一类是否包含0
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let (mut row0, mut column0) = (false, false);
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for i in 0..matrix.len() {
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for j in 0..matrix[0].len() {
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if matrix[i][j] == 0 {
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matrix[i][0] = 0;
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matrix[0][j] = 0;
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if i == 0 {
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row0 = true;
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}
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if j == 0 {
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column0 = true
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}
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}
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}
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}
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// 遍历赋值
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for i in 1..matrix.len() {
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for j in 1..matrix[i].len() {
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if matrix[i][0] == 0 || matrix[0][j] == 0 {
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matrix[i][j] = 0;
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}
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}
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}
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if row0 {
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for i in 0..matrix[0].len() {
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matrix[0][i] = 0;
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}
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}
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if column0 {
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for i in 0..matrix.len() {
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matrix[i][0] = 0;
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}
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}
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}
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}

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