139. 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.

Have you met this question in a real interview? 

Yes

Example

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

这个题目并不困难，需要计算一个前项和数组，然后将该数组排序，依次计算后项减前项的最小值，计算出分别计算出坐标，注意点，最小值的位置需要加一，因为最小值并不会被计算在内。

java

class Pairs {    int index;    int sum;    public Pairs(int index, int sum) {        this.index = index;        this.sum = sum;    }}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[] arr = new int[2];        if (nums == null || nums.length == 0) {            return arr;        }        if (nums.length == 1) {            arr[0] = arr[1] = 0;            return arr;        }        Pairs[] p = new Pairs[nums.length + 1];        p[0] = new Pairs(0, 0);        for (int i = 1; i <= nums.length; i++) {            p[i] = new Pairs(i, p[i - 1].sum + nums[i - 1]);        }        Comparator<Pairs> cmp = new Comparator() {            public int compare(Object a, Object b) {                return ((Pairs)a).sum - ((Pairs)b).sum;            }        };        Arrays.sort(p, cmp);        int max = Integer.MAX_VALUE;        for (int i = 1; i <= nums.length; i++) {            if (max > p[i].sum - p[i - 1].sum) {                max = p[i].sum - p[i - 1].sum;                arr[0] = p[i].index - 1;                arr[1] = p[i - 1].index - 1;                Arrays.sort(arr);            }        }        arr[0]++;        return arr;    }}