Cinte's Leetcode Record
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64. Minimum Path Sum

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Symbols count in article: 2k Reading time ≈ 2 mins.

64. Minimum Path Sum

动态规划
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class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        size_t m = grid.size(), n = grid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));
        dp[0][0] = grid[0][0];      
        for(size_t i = 0; i < m; ++i)
        {
            for(size_t j = 0; j < n; ++j)
            {
                if(i == 0 && j == 0)
                    continue;
                if(i == 0)
                    dp[i][j] = dp[i][j - 1] + grid[i][j];
                else if(j == 0)
                    dp[i][j] = dp[i - 1][j] + grid[i][j];
                else
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }
        return dp[m - 1][n - 1];
    }
};

T(n) : O(m * n)

S(n) : O(m * n)

尝试对S(n) 优化

观察代码，发现每次用到的数据只有dp[i][j], dp[i][j - 1]和dp[i - 1][j]，考虑dp[i][j], dp[i][j - 1]合并到一个一维数组，剩下的再生产一个一维数组。

参考unique path中的方法 

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class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        size_t m = grid.size(), n = grid[0].size();
        int* cur = new int[n];
        int* pre = new int[n];
        for (size_t i = 0; i < m; ++i)
        {
            for (size_t j = 0; j < n; ++j)
            {
                if (i == 0 && j == 0)
                    cur[j] = grid[i][j];
                else if (i == 0)
                    cur[j] = cur[j - 1] + grid[i][j];
                else if (j == 0)
                    cur[j] = pre[j] + grid[i][j];
                else
                    cur[j] = min(pre[j], cur[j - 1]) + grid[i][j];
            }
            swap(cur, pre);
        }
        return pre[n - 1];
    }
};
T(n) : O(m * n)

S(n) : O(2 * n)

再优化

pre[j] 与 cur[j] 可以合并

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class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        size_t m = grid.size(), n = grid[0].size();
        int* cur = new int[n];
        for (size_t i = 0; i < m; ++i)
        {
            for (size_t j = 0; j < n; ++j)
            {
                if (i == 0 && j == 0)
                    cur[j] = grid[i][j];
                else if (i == 0)
                    cur[j] = cur[j - 1] + grid[i][j];
                else if (j == 0)
                    cur[j] = cur[j] + grid[i][j];
                else
                    cur[j] = min(cur[j], cur[j - 1]) + grid[i][j];
            }
        }
        return cur[n - 1];
    }
};

T(n) : O(m * n)

S(n) : O(n)

review
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class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<int> dp(n, 0);
        dp[0] = grid[0][0];
        for(int i = 1; i < n; ++i)
            dp[i] = dp[i - 1] + grid[0][i];
        for(int i = 1; i < m; ++i)
            for(int j = 0; j < n; ++j)
                dp[j] = (j > 0 ? min(dp[j], dp[j - 1]) : dp[j]) + grid[i][j];
        return dp.back();
    }
};