【LeetCode】064. Minimum Path Sum

题目：

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.

题解：

Solution 1 ()

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = (int) grid.size(), n = (int) grid[].size();
        vector<long> dp(n,INT_MAX);
        dp[] = grid[][];
        for(int i=; i<m; ++i) {
            for(int j=; j<n; ++j) {
                if(j > )
                    dp[j] = min(dp[j] + grid[i][j], dp[j-] + grid[i][j]);
                else
                    if(i>) dp[j] = dp[j] + grid[i][j];
            }
        }
        return dp[n-];
    }
};

　　边界的第二种处理方法：因为Solution 1 中dp初始化为最大值，故需要考虑溢出情况，所以用long整型。这个就初始化为int整型。

Solution 2 ()

class Solution {
public:
    int minPathSum(vector<vector<int>>& nums) {
        int m = (int) nums.size();
        int n = (int) nums[].size();
        vector<int> v (n,);
        for (int i=; i<m; ++i) {
            for (int j=; j<n; ++j) {
                if (i>)
                    v[j] = nums[i][j] + ((j>) ? min(v[j], v[j-]) : v[j]);
                else
                    v[j] = nums[i][j] + ((j>) ? v[j-] : );
            }
        }
        return v[n-];
    }
};

　　解法没变，就是边界的处理上不一样，这个是先初始化边界了。

Solution 3 ()

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int dp[grid.size()][grid[].size()];
 
        dp[][] = grid[][];
        // init first row
        for(int i = ; i < grid[].size(); i ++){
            dp[][i] = dp[][i-] + grid[][i];
        }
        // init first col
        for(int i = ; i < grid.size(); i ++){
            dp[i][] = dp[i-][] + grid[i][];
        }
        for(int i = ; i < grid.size(); i ++){
            for(int j = ; j < grid[].size(); j++){
                dp[i][j] = dp[i - ][j] < dp[i][j-]? dp[i - ][j] + grid[i][j] : dp[i][j-] + grid[i][j];
            }
        }
        return dp[grid.size() - ][grid[].size() -];
    }
};