[LintCode] Subarray Sum Closest

Given an integer array, find a subarray with sum closest to zero. Return the indexes of the first number and last number.

Example

Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4]

Challenge

O(nlogn) time

http://www.lintcode.com/en/problem/subarray-sum-closest/

题目的意思是在一个数组中找一段连续的区间，使得这段区间的和的绝对值最小。做法就是利用前缀和，先用一个数组acc[i]来保存从nums[0]到nums[i]的和，同时还要记录下标，所以这里我用pair<int, int>来保存。那么，我们想要得到nums[i]到nums[j]的和，只要用acc[j] - acc[i-1]就可以了。但是这里有一点要注意要加一个辅助的节点，那就是[0, -1]，这样就可以确保可以找到以nums[0]开始的区间了。剩下的工作就是对acc数组排序，找到排序后相邻的差的绝对值最小的那一对节点。

 1 class Solution {

 2 public:

 3     /**

 4      * @param nums: A list of integers

 5      * @return: A list of integers includes the index of the first number 

 6      *          and the index of the last number

 7      */

 8     vector<int> subarraySumClosest(vector<int> nums){

 9         // write your code here

10         vector<pair<int, int> > acc;

11         acc.push_back(make_pair(0, -1));

12         int sum = 0;

13         for (int i = 0; i < nums.size(); ++i) {

14             sum += nums[i];

15             acc.push_back(make_pair(sum, i));

16         }

17         sort(acc.begin(), acc.end());

18         int min_abs = INT_MAX, a, b, tmp;

19         for (int i = 1; i < acc.size(); ++i) {

20             tmp = abs(acc[i].first - acc[i-1].first);

21             if (min_abs >= tmp) {

22                 min_abs = tmp;

23                 a = acc[i-1].second;

24                 b = acc[i].second;

25             }

26         }

27         vector<int> res;

28         res.push_back(min(a, b) + 1);

29         res.push_back(max(a, b));

30         return res;

31     }

32 };