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MinimumPathSum.java
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62 lines (40 loc) · 1.26 KB
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package recursion_and_dynamic_programming;
/**
* @Author: Wenhang Chen
* @Description: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。
* <p>
* 说明:每次只能向下或者向右移动一步。
* <p>
* 示例:
* <p>
* 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。
* @Date: Created in 8:36 3/11/2020
* @Modified by:
*/
public class MinimumPathSum {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length < 1) return 0;
// dp[i][j]存储当前位置最小花费
int[][] dp = new int[grid.length][grid[0].length];
dp[0][0] = grid[0][0];
for (int i = 1; i < grid.length; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int i = 1; i < grid[0].length; i++) {
dp[0][i] = dp[0][i - 1] + grid[0][i];
}
for (int i = 1; i < grid.length; i++) {
for (int j = 1; j < grid[0].length; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[grid.length - 1][grid[0].length - 1];
}
}