LeetCode 64. Minimum Path Sum

原题链接在这里：https://leetcode.com/problems/minimum-path-sum/

题目：

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.

题解：

与Unique Paths相似。存储历史信息是走到走到历史点的最小权重路径，更新当前点用上面点和左面点中的小值加上grid中当前点的值。

也相同的第二种降维方法.

Time Complexity: O(m*n). Space: O(n). 一维dp优化.

AC Java:

public class Solution {
    public int minPathSum(int[][] grid) {
        /*
        //Method 1
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        int m = grid.length;
        int n = grid[0].length;
        int [][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        for(int i = 1; i<m; i++){
            dp[i][0] = dp[i-1][0] + grid[i][0];
        }
        for(int j = 1; j<n; j++){
            dp[0][j] = dp[0][j-1] + grid[0][j];
        }
        for(int i = 1; i<m; i++){
            for(int j = 1; j<n; j++){
                dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
            }
        }
        return dp[m-1][n-1];
        */
        
        //Method 2
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        int m = grid.length;
        int n = grid[0].length;
        int [] dp = new int[n];
        dp[0] = grid[0][0];
        for(int j = 1; j<n; j++){
            dp[j] = dp[j-1] + grid[0][j];
        }
        for(int i = 1; i<m; i++){
            for(int j = 0; j<n; j++){
                if(j == 0){
                    dp[j] += grid[i][j];
                }else{
                    dp[j] = Math.min(dp[j],dp[j-1]) + grid[i][j];
                }
                
            }
        }
        return dp[n-1];
    }
}