題目描述:
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.
例子:
[[1,3,1], [1,5,1], [4,2,1]]Given the above grid map, return
7
. Because the path 1→3→1→1→1 minimizes the sum.分析:
題意:給定一個m x n 二維地圖(每個元素代表一個非負整型數),左上角爲起點、右下角爲終點,只能往下、往右走。返回最小路徑和的值。
思路:這道題採用動態規劃的方法:我們構造數組dp[m][n],dp[i][j]表示從位置(0, 0)到達位置(i, j)的最小路徑和。初始化各元素值爲0(其中dp[0][0] = 1,dp[i][0] = dp[i - 1][0] + grid[i][0](i∈1→m - 1),dp[0][j] = dp[0][j - 1] + grid[0][j](j∈1→n - 1)),那麼對於位置(i, j),根據題意,只能從其上方或者左側移動過來兩種可能,我們選擇較小值。因此狀態轉移方程爲:
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]。
時間複雜度爲O(m * n)。
代碼:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
// Exceptional Case:
if(grid.empty()){
return 0;
}
int m = grid.size(), n = grid[0].size();
// create
vector<vector<int>> dp(m, vector<int>(n, 0));
// init
dp[0][0] = grid[0][0];
for(int i = 1; i <= n - 1; i++){
dp[0][i] = dp[0][i - 1] + grid[0][i];
}
for(int i = 1; i <= m - 1; i++){
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
// dp
for(int i = 1; i <= m - 1; i++){
for(int j = 1; j <= n - 1; j++){
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
};