Triangle(LeetCode)
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题目:
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).
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:
刚开始,把这道题想复杂了。搞了好多递归什么的,一直都TimeLimitedExceed。原因是刚开始思路一直是从上往下。其实只要从下往上思考就很简单。
- 用temp[ ]复制最后一行的值。
- 对于倒数第二行的第i个值,从temp[ ]中的第i个值和第i+1个值(因为这两个是倒数第二行第i个值,在最后一行的两个相邻值)中选择小的那个加到temp[i]上。
- 类似的,再把temp上值选择性地加到倒数第三行上,并记录在temp上。重复直到加到第一行。
- 返回temp[0]
代码:
public class Solution { public int minimumTotal(List<List<Integer>> triangle) { int n = triangle.size();//n represents the nth line int[] temp = new int[n]; for (int i = 0; i < n; i++){//copy the last line to temp[] temp[i] = triangle.get(n - 1).get(i); } n--; while (n > 0){ for (int i = 0; i < n; i++){ if (temp[i] <= temp[i+1]){ temp[i] += triangle.get(n-1).get(i); }else{ temp[i] = temp[i+1] + triangle.get(n-1).get(i); } } n--; } return temp[0]; }}
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