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  • 64. Minimum Path Sum

    #week13

    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 1:

    [[1,3,1],
     [1,5,1],
     [4,2,1]]
    

    Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

    分析:

    这个动态规划的状态转换很明显:

    sum[i][j]  = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];

    要么从上面走过来要么从左边走过来,才能够得到最小的,因为边的权值为正数,所以不可能是右边和下面

    然后最后的结果就是sum[m - 1][n - 1]

    题解:

     1 class Solution {
     2 public:
     3     int minPathSum(vector<vector<int>>& grid) {
     4         int m = grid.size();
     5         int n = grid[0].size(); 
     6         vector<vector<int> > sum(m, vector<int>(n, grid[0][0]));
     7         for (int i = 1; i < m; i++)
     8             sum[i][0] = sum[i - 1][0] + grid[i][0];
     9         for (int j = 1; j < n; j++)
    10             sum[0][j] = sum[0][j - 1] + grid[0][j];
    11         for (int i = 1; i < m; i++)
    12             for (int j = 1; j < n; j++)
    13                 sum[i][j]  = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
    14         return sum[m - 1][n - 1];
    15     }
    16 };
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  • 原文地址:https://www.cnblogs.com/iamxiaoyubei/p/8278272.html
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