leetcode——[064]Minimum Path Sum最短路径和

题目

给定一个包含非负整数的 m x n 网格，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。

说明：每次只能向下或者向右移动一步。

示例:

输入:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
输出: 7
解释: 因为路径 1→3→1→1→1 的总和最小。

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.

解题方法

简单动态规划问题，状态转移方程为res[i][j] = min(res[i-1][j], res[i][j-1]) + grid[i][j]，其中res保存到达每个格子的最短路径和，grid保存每个格子的值，到grid[i][j]的最短路径和为到其左边或上边的最短路径和中的最小值加grid[i][j]。这段代码跑了5ms，超过了93.86%的Java提交。

public class Solution {
    // Method 1 5ms 93.86%
    public int minPathSum(int[][] grid) {
        int m = grid.length;
        int x = grid[0].length;
        int[][] res = new int[m][x];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < x; j++) {
                if (i == 0 && j == 0) {
                    res[i][j] = grid[i][j];
                }
                else if (i == 0) {
                    res[i][j] = res[i][j-1] + grid[i][j];
                }
                else if (j == 0) {
                    res[i][j] = res[i-1][j] + grid[i][j];
                }
                else{
                    res[i][j] = Math.min(res[i-1][j], res[i][j-1]) + grid[i][j];
                }
            }
        }
        return res[m-1][x-1];
    }
}