LeetCode Minimum Path Sum

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

是DP题，与Unique Paths 和 Unique Paths II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。

跟Unique Paths 和 Unique Paths II  都有相同的第二种降维方法。

AC Java:

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