Continuous Subarray Sum

最新推荐文章于 2025-07-29 14:24:32 发布

转载最新推荐文章于 2025-07-29 14:24:32 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/beiyeqingteng/p/5631733.html

本文介绍了一种高效算法来解决最大连续子数组求和的问题，即找出整数数组中连续子数组的最大和，并返回该子数组的起始与结束索引。通过动态规划的方法，实现了O(n)的时间复杂度。

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Given an integer array, find a continuous subarray where the sum of numbers is the biggest. Your code should return the index of the first number and the index of the last number. (If their are duplicate answer, return anyone)

Example

Give A = [-3, 1, 3, -3, 4], return [1,4].

分析：

max(i) = {max(i - 1) + A[i] if max(i - 1) >= 0, otherwise, A[i]}

 1 public class Solution {
 2     /**
 3      * @param A an integer array
 4      * @return  A list of integers includes the index of the first number and the index of the last number
 5      * cnblogs.com/beiyeqingteng/
 6      */
 7     public ArrayList<Integer> continuousSubarraySum(int[] A) {
 8         if (A == null || A.length == 0) return null;
 9         
10         ArrayList<Integer> list = new ArrayList<Integer>();
11         // initialization
12         int preMax = A[0];
13         int startIndex = 0;
14         int tempMax = A[0];
15         list.add(0);
16         list.add(0);
17         
18         for (int i = 1; i < A.length; i++) {
19             if (preMax < 0) {
20                 preMax = A[i];
21                 startIndex = i;
22             } else {
23                 preMax = A[i] + preMax;
24             }
25             // update
26             if (preMax > tempMax) {
27                 list.clear();
28                 list.add(startIndex);
29                 list.add(i);
30                 tempMax = preMax;
31             }
32         }
33         return list;
34     } 
35 }

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转载于:https://www.cnblogs.com/beiyeqingteng/p/5631733.html