[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。

class Solution {
public:
/**
* @param nums: A list of integers
* @return: A list of integers includes the index of the first number
*          and the index of the last number
*/
vector<int> subarraySumClosest(vector<int> nums){
// write your code here
vector<pair<int, int> > acc;
acc.push_back(make_pair(0, -1));
int sum = 0;
for (int i = 0; i < nums.size(); ++i) {
sum += nums[i];
acc.push_back(make_pair(sum, i));
}
sort(acc.begin(), acc.end());
int min_abs = INT_MAX, a, b, tmp;
for (int i = 1; i < acc.size(); ++i) {
tmp = abs(acc[i].first - acc[i-1].first);
if (min_abs >= tmp) {
min_abs = tmp;
a = acc[i-1].second;
b = acc[i].second;
}
}
vector<int> res;
res.push_back(min(a, b) + 1);
res.push_back(max(a, b));
return res;
}
};