[leetcode]64Minimum Path Sum 动态规划

/**
 * 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.
 */
/*
* 动态规划题目,思路是把每一个点的最小和求出来,最后返回右下角的点。
* 每个点的最小和就是比较它左边和上边的点,把较小的那个和本身加起来
* 首先应该吧最左边列和最上边一行的点的最小和求出,因为后边要用到,且他们的最小和求法和中间的不一样
* 难点:初始值很多,是左边一列和上边一行所有的点,都可以直接求出*/
public class Q64MinimumPathSum {
    public static void main(String[] args) {

    }
    public int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        //求出最左边一行的每个点对应的最小和
        for (int i = 1; i < m; i++) {
            grid[i][0] += grid[i-1][0];
        }
        //最上边行的
        for (int i = 1; i < n; i++) {
            grid[0][i] += grid[0][i-1];
        }
        //前两个是判断特殊情况
        if (m == 1)
            return grid[0][n-1];
        else if (n == 1)
            return grid[m-1][0];
        //动态规划主体
        else
        {
            int temp;
            for (int i = 1; i < m; i++) {
                for (int j = 1; j < n; j++) {
                    //求该点的最小和
                    temp = Math.min(grid[i-1][j],grid[i][j-1]);
                    grid[i][j] += temp;
                }
            }
            //右下角的点
            return grid[m-1][n-1];
        }
    }
}
原文地址:https://www.cnblogs.com/stAr-1/p/7251199.html