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];
    }
};
原文地址:https://www.cnblogs.com/zhonghuasong/p/7761857.html