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LeetCode Triangle
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 is11(i.e., 2 + 3 + 5 + 1 = 11).
Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
思路:从下往上,每一行的结果根据下面一行的路基累计和而计算。(参考大神才晓得)
triangle[i][j] += min(triangle[i + 1][j], triangle[i + 1][j + 1])
对于此题子在面试的时候, 除了题意,还要要问清楚这些题的其他问题,比如 :传递的数据元素可以修改否? 最后求出的和sum会不会超出integer的范围?
package cn.edu.algorithm.huawei;
import java.util.ArrayList;
public class Solution {
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return -1;
}
int len = triangle.size();
int[][] dp = new int[len][len];
for (int i = 0; i < len; i++) {
dp[len - 1][i] = triangle.get(len - 1).get(i);
}
for (int i = len - 2; i >= 0; i--) {
for (int j = 0; j <= i; j++) {
dp[i][j] = Math.min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle.get(i).get(j);
}
}
return dp[0][0];
}
public static void main(String[] args) {
ArrayList<ArrayList<Integer>> triangle = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> list1 = new ArrayList();
list1.add(1);
ArrayList<Integer> list2 = new ArrayList();
list2.add(2);
list2.add(3);
triangle.add(list1);
triangle.add(list2);
Solution s = new Solution();
int path = s.minimumTotal(triangle);
System.out.println(path);
}
}
LeetCode Triangle
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