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[LeetCode] 64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

最小路径和。题意是给一个二维矩阵，每个格子上都有一个非负整数。请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。

这是动态规划的入门题。DP[i][j]的含义是代表走到坐标(i, j)位置的最小路径和。初始状态dp[0][0] = grid[0][0]，因为对于某一个坐标来说，只会是从上面一个单元格或者是左边的一个单元格过来，所以转移方程是dp[i][j] = min(dp[i−1][j], dp[i][j−1]) + grid[i][j]。

时间O(mn)

空间O(1)

Java实现

 1 class Solution {
 2     public int minPathSum(int[][] grid) {
 3         int m = grid.length;
 4         int n = grid[0].length;
 5         int[][] dp = new int[m][n];
 6         dp[0][0] = grid[0][0];
 7         for (int i = 1; i < m; i++) {
 8             dp[i][0] = dp[i - 1][0] + grid[i][0];
 9         }
10         
11         for (int j = 1; j < n; j++) {
12             dp[0][j] = dp[0][j - 1] + grid[0][j];
13         }
14         
15         for (int i = 1; i < m; i++) {
16             for (int j = 1; j < n; j++) {
17                 dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
18             }
19         }
20         return dp[m - 1][n - 1];
21     }
22 }

JavaScript实现

 1 /**
 2  * @param {number[][]} grid
 3  * @return {number}
 4  */
 5 var minPathSum = function (grid) {
 6     for (let i = 0; i < grid.length; i++) {
 7         for (let j = 0; j < grid[0].length; j++) {
 8             if (i == 0 && j != 0) {
 9                 grid[i][j] += grid[i][j - 1];
10             }
11             if (j == 0 && i != 0) {
12                 grid[i][j] += grid[i - 1][j];
13             }
14             if (i != 0 && j != 0) {
15                 grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
16             }
17         }
18     }
19     return grid[grid.length - 1][grid[0].length - 1];
20 };