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.

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Example

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

这道题需要新建一个class记录前i个数的sum和index。注意要新建一个pair(0,-1)来表示index 0之前的和等于0。遍历整个nums，计算前i个数的和，之后排序pairs数组，找到相邻差值最小的一对，保存下来index，这里需要注意，即使是相邻，他们的记录的下标可能是一前一后，或者以后一前，在赋值给res之前，先赋值给temp数组，对temp数组排序保证前小后大，在赋值给res。代码如下：

public class Solution {    /**     * @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     */    class Pair {        int sum, index;        public Pair(int _sum, int _index) {            this.sum = _sum;            this.index = _index;        }    }        public int[] subarraySumClosest(int[] nums) {        // write your code here        Pair[] pairs = new Pair[nums.length + 1];        pairs[0] = new Pair(0, -1);        for (int i = 1; i <= nums.length; i ++) {            pairs[i] = new Pair(pairs[i - 1].sum + nums[i - 1], i - 1);        }        Arrays.sort(pairs, new Comparator<Pair>(){            public int compare(Pair p1, Pair p2) {                return p1.sum - p2.sum;            }        });        int min = Integer.MAX_VALUE;        int[] res = new int[2];        for (int i = 1; i < pairs.length; i ++) {            if (min > pairs[i].sum - pairs[i - 1].sum) {                min = pairs[i].sum - pairs[i - 1].sum;                int[] temp = new int[2];                temp[0] = pairs[i].index;                temp[1] = pairs[i - 1].index;                Arrays.sort(temp);                res[0] = temp[0] + 1;                res[1] = temp[1];            }        }        return res;    }}