lintcode 容易题：Minimum Path Sum 最小路径和

题目：

最小路径和

给定一个只含非负整数的m*n网格，找到一条从左上角到右下角的可以使数字和最小的路径。

样例

注意

你在同一时间只能向下或者向右移动一步

解题：

这个和求三角形的最小路径的差不多，这里是个矩阵，第一列和第一行要单独处理，每一点的值等于自身的值加上上一点的值，对于中间节点：grid[i][j] + = min( grid[i-1][j] , grid[i][j-1]) ,也就是说，i j 点的值只能是其左侧的点或者上侧的点加过来，这个时间复杂度比较高O(N²)

Java程序：

public class Solution {
    /**
     * @param grid: a list of lists of integers.
     * @return: An integer, minimizes the sum of all numbers along its path
     */
    public int minPathSum(int[][] grid) {
        // write your code here
        int m = grid.length;
        if(m == 0 )
            return 0;
        int n = grid[0].length;
        if( n==0 )
            return 0;
        for(int i=1 ; i<n; i++ ){
            grid[0][i] += grid[0][i-1];
        }
        for( int i= 1;i< m; i++ ){
            grid[i][0] += grid[i-1][0];
        }
        for(int i=1; i< m ; i++){
            for(int j = 1;j< n; j++){
                grid[i][j] += Math.min(grid[i-1][j],grid[i][j-1]);
            }
        }
        return grid[m-1][n-1];
    }
}

Python程序：

class Solution:
    """
    @param grid: a list of lists of integers.
    @return: An integer, minimizes the sum of all numbers along its path
    """
    def minPathSum(self, grid):
        # write your code here
        m = len(grid)
        if m == 0:
            return 0
        n = len(grid[0])
        if n==0:
            return 0
        for i in range(m):
            for j in range(n):
                if i==0 and j!=0:
                    grid[i][j] += grid[i][j-1]
                elif j==0 and i!=0:
                    grid[i][j] += grid[i-1][j]
                elif j!=0 and i!=0:
                    grid[i][j] += min(grid[i-1][j] , grid[i][j-1])
        return grid[m-1][n-1]