[LintCode]Subarray Sum Closest

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

找到和最接近0的子数组

前缀和，并且记录坐标，然后对于前缀和排序，找出相邻两个之差值最接近0的，差值肯定是正数，但是index前后位置关系不定，所以子数组的和可正可负

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     */    public int[] subarraySumClosest(int[] nums) {        // write your code here        int[] res = new int[2];        if (nums == null || nums.length <= 1) {            return res;        }        int len = nums.length;        Pair[] pairs = new Pair[len + 1];        pairs[0] = new Pair(0, 0);        for (int i = 0; i < len; i++) {            // 因为子数组0 ~ i可能和为0，最开始位置和为0，这样后续排序的时候会有影响，所以index要加一            pairs[i + 1] = new Pair(pairs[i].sum + nums[i], 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;        for (int i = 1; i <= len; i++) {            if (min > pairs[i].sum - pairs[i - 1].sum) {                int[] temp = {pairs[i].index - 1, pairs[i - 1].index - 1};                min = pairs[i].sum - pairs[i - 1].sum;                Arrays.sort(temp);                res[0] = temp[0] + 1;                res[1] = temp[1];            }        }        return res;    }}class Pair {    int sum;    int index;    public Pair(int sum, int index) {        this.sum = sum;        this.index = index;    }}

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