leetcode 120.Triangle | Java最短代码实现

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]

The minimum path sum from top to bottom is

11

(i.e., 2 + 3 + 5 + 1 =
11).

【抛砖】

考察动态规划，这里从底层开始，逐层往上层动态求和，最后得到的最小和就是dp[0]：

public int minimumTotal(List<List<Integer>> triangle) {
int[] dp = new int[triangle.size() + 1];
for (int i = triangle.size() - 1; i >= 0; i--)
for (int j = 0; j < triangle.get(i).size(); j++)
dp[j] = triangle.get(i).get(j) + Math.min(dp[j], dp[j + 1]);
return dp[0];
}

43 / 43 test
cases passed.
Runtime: 6 ms Your
runtime beats 40.91% of javasubmissions.

【补充】

同样的动归，采用深度优先，很容易想到用递归法，但是这种方法在数据量很大时超时了：

public int minimumTotal(List<List<Integer>> triangle) {
return findMinPath(triangle, 0, Integer.MAX_VALUE, 0, 0);
}
public int findMinPath(List<List<Integer>> triangle, int curSum, int min, int index, int level) {
curSum += triangle.get(level).get(index);
if (level == triangle.size() - 1)
return Math.min(min, curSum);
return Math.min(findMinPath(triangle, curSum, min, index, level + 1),
findMinPath(triangle, curSum, min, index + 1, level + 1));
}

欢迎优化！