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  • LeetCode——minimum-path-sum

    Question

    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.

    Solution

    1. 动态规划。 p[i][j] = grid[i][j] + min(p[i - 1][j], p[i][j - 1]). 因为到达一个节点,最多有两种选择,当然是选择代价较小的。

    2. 时间复杂度O(n^2)。

    Code

    class Solution {
    public:
        int minPathSum(vector<vector<int> > &grid) {
            if (grid.size() <= 0)
                return 0;
            vector<vector<int>> tb(grid.size(), vector<int>(grid[0].size(), 0));
            int row = grid.size();
            int col = grid[0].size();
            tb[0][0] = grid[0][0];
            // 注意初始化是一个代价累加的过程
            for (int i = 1; i < row; i++) {
                tb[i][0] = grid[i][0] + tb[i - 1][0];
            }
            for (int i = 1; i < col; i++) {
                tb[0][i] = grid[0][i] + tb[0][i - 1];
            }
            for (int i = 1; i < row; i++) {
                for (int j = 1; j < col; j++) {
                    tb[i][j] = grid[i][j] + min(tb[i - 1][j], tb[i][j - 1]);
                }
            }
            return tb[row - 1][col - 1];
        }
    };
    
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  • 原文地址:https://www.cnblogs.com/zhonghuasong/p/7761857.html
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