Minimum Path Sum
最后更新于：2022-04-02 01:12:48
# Minimum Path Sum
- tags: [[DP_Matrix](# "根据动态规划解题的四要素，矩阵类动态规划问题通常可用 f[x][y] 表示从起点走到坐标(x,y)的值")]
### Source
- lintcode: [(110) Minimum Path Sum](http://www.lintcode.com/en/problem/minimum-path-sum/)
~~~
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
1. State: f[x][y] 从坐标(0,0)走到(x,y)的最短路径和
1. Function: f[x][y] = (x, y) + min{f[x-1][y], f[x][y-1]}
1. Initialization: f[0][0] = A[0][0], f[i][0] = sum(0,0 -> i,0), f[0][i] = sum(0,0 -> 0,i)
1. Answer: f[m-1][n-1]
注意最后返回为f[m-1][n-1]而不是f[m][n].
首先看看如下正确但不合适的答案，OJ上会出现[TLE](# "Time Limit Exceeded 的简称。你的程序在 OJ 上的运行时间太长了，超过了对应题目的时间限制。")。未使用hashmap并且使用了递归的错误版本。
### C++ [dfs](# "Depth-First Search, 深度优先搜索") without hashmap: ~~Wrong answer~~
~~~
class Solution {
public:
/**
* @param grid: a list of lists of integers.
* @return: An integer, minimizes the sum of all numbers along its path
*/
int minPathSum(vector > &grid) {
if (grid.empty()) {
return 0;
}
const int m = grid.size() - 1;
const int n = grid[0].size() - 1;
return helper(grid, m, n);
}
private:
int helper(vector > &grid_in, int x, int y) {
if (0 == x && 0 == y) {
return grid_in[0][0];
}
if (0 == x) {
return helper(grid_in, x, y - 1) + grid_in[x][y];
}
if (0 == y) {
return helper(grid_in, x - 1, y) + grid_in[x][y];
}
return grid_in[x][y] + min(helper(grid_in, x - 1, y), helper(grid_in, x, y - 1));
}
};
~~~
使用迭代思想进行求解的正确实现：
### C++ Iterative
~~~
class Solution {
public:
/**
* @param grid: a list of lists of integers.
* @return: An integer, minimizes the sum of all numbers along its path
*/
int minPathSum(vector > &grid) {
if (grid.empty() || grid[0].empty()) {
return 0;
}
const int M = grid.size();
const int N = grid[0].size();
vector > ret(M, vector (N, 0));
ret[0][0] = grid[0][0];
for (int i = 1; i != M; ++i) {
ret[i][0] = grid[i][0] + ret[i - 1][0];
}
for (int i = 1; i != N; ++i) {
ret[0][i] = grid[0][i] + ret[0][i - 1];
}
for (int i = 1; i != M; ++i) {
for (int j = 1; j != N; ++j) {
ret[i][j] = grid[i][j] + min(ret[i - 1][j], ret[i][j - 1]);
}
}
return ret[M - 1][N - 1];
}
};
~~~
### 源码分析
1. 异常处理，不仅要对grid还要对grid[0]分析
1. 对返回结果矩阵进行初始化，注意ret[0][0]须单独初始化以便使用ret[i-1]
1. 递推时i和j均从1开始
1. 返回结果ret[M-1][N-1]，注意下标是从0开始的
此题还可进行空间复杂度优化，和背包问题类似，使用一维数组代替二维矩阵也行，具体代码可参考 [水中的鱼: [LeetCode] Minimum Path Sum 解题报告](http://fisherlei.blogspot.sg/2012/12/leetcode-minimum-path-sum.html)
优化空间复杂度，要么对行遍历进行优化，要么对列遍历进行优化，通常我们习惯先按行遍历再按列遍历，有状态转移方程 f[x][y] = (x, y) + min{f[x-1][y], f[x][y-1]} 知，想要优化行遍历，那么f[y]保存的值应为第x行第y列的和。由于无行下标信息，故初始化时仅能对第一个元素初始化，分析时建议画图理解。
### C++ 1D vector
~~~
class Solution {
public:
/**
* @param grid: a list of lists of integers.
* @return: An integer, minimizes the sum of all numbers along its path
*/
int minPathSum(vector > &grid) {
if (grid.empty() || grid[0].empty()) {
return 0;
}
const int M = grid.size();
const int N = grid[0].size();
vector ret(N, INT_MAX);
ret[0] = 0;
for (int i = 0; i != M; ++i) {
ret[0] = ret[0] + grid[i][0];
for (int j = 1; j != N; ++j) {
ret[j] = grid[i][j] + min(ret[j], ret[j - 1]);
}
}
return ret[N - 1];
}
};
~~~
初始化时需要设置为`INT_MAX`，便于`i = 0`时取`ret[j]`.
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