Lintcode: Subarray Sum

Given an integer array, find a subarray where the sum of numbers is zero. Your code should return the index of the first number and the index of the last number.

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

推荐解法：The idea is based on the prefix sum: Iterate through the array and for every element array【i】, calculate sum of elements form 0 to i (this can simply be done as sum += arr【i】). If the current sum has been seen before, then there is a zero sum array, the start and end index are returned.

用HashMap: O(N)时间，但是more memory, 大case会MLE

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 ArrayList<Integer> subarraySum(int[] nums) {
        // write your code here
        ArrayList<Integer> res = new ArrayList<Integer>();
        if (nums==null || nums.length==0) return res;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, -1);
        int sum = 0;
        for (int i=0; i<nums.length; i++) {
            sum += nums[i];
            if (!map.containsKey(sum)) {
                map.put(sum, i);
            }
            else {
                res.add(map.get(sum)+1);
                res.add(i);
                return res;
            }
        }
        return res;
    }
}

因为上面这个简洁的代码会MLE，所以(nlog(n))第二个算法，时间多一点，但是空间少一点

class Element implements Comparable<Element>{
    int index;
    int value;
    public Element(int i, int v){
        index = i;
        value = v;
    }
    public int compareTo(Element other){
        return this.value-other.value;
    }
    public int getIndex(){
        return index;
    }
    public int getValue(){
        return value;
    }
}

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 ArrayList<Integer> subarraySum(int[] nums) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if (nums==null || nums.length==0) return res;
        int len = nums.length;
        Element[] sums = new Element[len+1];
        sums[0] = new Element(-1,0);
        int sum = 0;
        for (int i=0;i<len;i++){
            sum += nums[i];
            sums[i+1] = new Element(i,sum);
        }
        Arrays.sort(sums);
        for (int i=0;i<len;i++)
            if (sums[i].getValue()==sums[i+1].getValue()){
                int start = Math.min(sums[i].getIndex(),sums[i+1].getIndex())+1;
                int end = Math.max(sums[i].getIndex(),sums[i+1].getIndex());
                res.add(start);
                res.add(end);
                return res;
            }

        return res;
    }
}