Leetcode64. 最小路径和
题目:
给定一个包含非负整数的 m x n 网格,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小。
说明:每次只能向下或者向右移动一步。
示例:
输入:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
输出: 7
解释: 因为路径 1→3→1→1→1 的总和最小。
题解:
- 当左边和上边都不是矩阵边界时: 即当i≠0,j≠0i \not= 0, j \not= 0i=0,j=0时,dp[i][j]=min(dp[i−1][j],dp[i][j−1])+grid[i][j]dp[i][j] = min(dp[i-1][j], dp[i][j - 1]) + grid[i][j]dp[i][j]=min(dp[i−1][j],dp[i][j−1])+grid[i][j];
- 当只有左边是矩阵边界时: 只能从上面来,即当i=0,j≠0i = 0, j \not= 0i=0,j=0 时, dp[i][j]=dp[i][j−1]+grid[i][j]dp[i][j] = dp[i][j-1] +grid[i][j]dp[i][j]=dp[i][j−1]+grid[i][j] ;
- 当只有上边是矩阵边界时: 只能从左面来,即当i≠0,j=0i \not= 0, j = 0i=0,j=0 时, dp[i][j]=dp[i−1][j]+grid[i][j]dp[i][j] = dp[i - 1][j]+ grid[i][j]dp[i][j]=dp[i−1][j]+grid[i][j];
- 当左边和上边都是矩阵边界时: 即当i=0,j=0i = 0, j = 0i=0,j=0时,其实就是起点, dp[i][j]=grid[i][j]dp[i][j] = grid[i][j]dp[i][j]=grid[i][j]。
java代码:
/**
* @param grid
* @return
*/
public static int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) continue;
else if (i == 0) grid[i][j] = grid[i][j - 1] + grid[i][j];
else if (j == 0) grid[i][j] = grid[i - 1][j] + grid[i][j];
else grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j]) + grid[i][j];
}
}
return grid[m-1][n-1];
}