Problem: 64. 最小路径和

思路

👨‍🏫 路飞

💖 朴素版

⏰ 时间复杂度: $O(nm)$
🌎 空间复杂度: $O(nm)$

class Solution {public int minPathSum(int[][] grid){int n = grid.length;if (n == 0)return 0;int m = grid[0].length;int[][] f = new int[n][m];f[0][0] = grid[0][0];for (int i = 1; i < m; i++)f[0][i] = f[0][i - 1] + grid[0][i];for (int i = 1; i < n; i++)f[i][0] = f[i - 1][0] + grid[i][0];for (int i = 1; i < n; i++)for (int j = 1; j < m; j++)f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];return f[n - 1][m - 1];}
}

💖 空间优化版

⏰ 时间复杂度: $O(nm)$
🌎 空间复杂度: $O(1)$

class Solution {public int minPathSum(int[][] grid) {for(int i = 0; i < grid.length; i++) {for(int j = 0; j < grid[0].length; j++) {if(i == 0 && j == 0) continue;//不处理else if(i == 0)  grid[i][j] = grid[i][j - 1] + grid[i][j];//第一行的，只能从左边过来else if(j == 0)  grid[i][j] = grid[i - 1][j] + grid[i][j];//第一列的，只能从上面过来else grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j];//从上面或者左边较小的地方过来}}return grid[grid.length - 1][grid[0].length - 1];}
}