题目链接
题目描述
实现代码
动态规划思想:记dp[i][j]为从起点(0, 0)到(i,j)的最小路径和,首先初始化条件:
- dp[0][0] = grid[0][0];
- dp[0][j] = grid[0][j] + dp[0][j-1];
- dp[i][0] = grid[i][0] + dp[i-1][0];
动态转移方程为:
dp[i][j] = Math.max(dp[i-1][j]. dp[i][j-1]) + grid[i][j]
实现代码:
class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = grid[0][0];
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
}
在二维数组动态规划的基础上进行降维优化,实现代码:
class Solution {
public int minPathSum(int[][] grid) {
int n = grid.length, m = grid[0].length;
int[] f = grid[0];
for(int i = 1; i < m; i++) f[i] += f[i - 1];
for(int i = 1; i < n; i++){
for(int j = 0 ; j < m; j++){
if(j > 0) f[j] = Math.min(f[j], f[j - 1]) + grid[i][j];
else f[j] += grid[i][j];
}
}
return f[m - 1];
}
}