Minimum Path Sum

描述

Given a m × 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

分析

跟第 ??? 节 Unique Paths 很类似。

设状态为f[i][j],表示从起点(0,0)到达(i,j)的最小路径和,则状态转移方程为:

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

备忘录法

// Minimum Path Sum
// 备忘录法
class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        const int m = grid.size();
        const int n = grid[0].size();
        this->f = vector<vector<int> >(m, vector<int>(n, -1));
        return dfs(grid, m-1, n-1);
    }
private:
    vector<vector<int> > f;  // 缓存

    int dfs(const vector<vector<int> > &grid, int x, int y) {
        if (x < 0 || y < 0) return INT_MAX; // 越界,终止条件,注意,不是0

        if (x == 0 && y == 0) return grid[0][0]; // 回到起点,收敛条件

        return min(getOrUpdate(grid, x - 1, y),
                getOrUpdate(grid, x, y - 1)) + grid[x][y];
    }

    int getOrUpdate(const vector<vector<int> > &grid, int x, int y) {
        if (x < 0 || y < 0) return INT_MAX; // 越界,注意,不是0
        if (f[x][y] >= 0) return f[x][y];
        else return f[x][y] = dfs(grid, x, y);
    }
};

动规

// Minimum Path Sum
// 二维动规
class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        if (grid.size() == 0) return 0;
        const int m = grid.size();
        const int n = grid[0].size();

        int f[m][n];
        f[0][0] = grid[0][0];
        for (int i = 1; i < m; i++) {
            f[i][0] = f[i - 1][0] + grid[i][0];
        }
        for (int i = 1; i < n; i++) {
            f[0][i] = f[0][i - 1] + grid[0][i];
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                f[i][j] = min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
            }
        }
        return f[m - 1][n - 1];
    }
};

动规+滚动数组

// Minimum Path Sum
// 二维动规+滚动数组
class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        const int m = grid.size();
        const int n = grid[0].size();

        int f[n];
        fill(f, f+n, INT_MAX); // 初始值是 INT_MAX,因为后面用了min函数。
        f[0] = 0;

        for (int i = 0; i < m; i++) {
            f[0] += grid[i][0];
            for (int j = 1; j < n; j++) {
                // 左边的f[j],表示更新后的f[j],与公式中的f[i[[j]对应
                // 右边的f[j],表示老的f[j],与公式中的f[i-1][j]对应
                f[j] = min(f[j - 1], f[j]) + grid[i][j];
            }
        }
        return f[n - 1];
    }
};

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