Minimum Path Sum问题及解法
2017-07-29 20:35
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问题描述:
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.
问题分析:
分析同Unique Paths。构造状态转移的dp数组,确定关系dp[i][j]
= min(dp[i - 1][j],dp[i][j - 1]) + grid[i - 1][j - 1],最终得到答案dp[m]
。
过程详见代码:class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MAX));
dp[0][1] = dp[1][0] = 0;
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i - 1][j - 1];
}
}
return dp[m]
;
}
};
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.
问题分析:
分析同Unique Paths。构造状态转移的dp数组,确定关系dp[i][j]
= min(dp[i - 1][j],dp[i][j - 1]) + grid[i - 1][j - 1],最终得到答案dp[m]
。
过程详见代码:class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MAX));
dp[0][1] = dp[1][0] = 0;
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i - 1][j - 1];
}
}
return dp[m]
;
}
};
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