[LintCode]Subarray Sum Closest

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

找到和最接近0的子數組

前綴和，並且記錄座標，然後對於前綴和排序，找出相鄰兩個之差值最接近0的，差值肯定是正數，但是index前後位置關係不定，所以子數組的和可正可負

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++) {
            // 因爲子數組0 ~ i可能和爲0，最開始位置和爲0，這樣後續排序的時候會有影響，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;
    }
}