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

 1 public class Solution {

 2     /**

 3      * @param nums: A list of integers

 4      * @return: A list of integers includes the index of the first number 

 5      *          and the index of the last number

 6      */

 7     public ArrayList<Integer> subarraySum(int[] nums) {

 8         // write your code here

 9         ArrayList<Integer> res = new ArrayList<Integer>();

10         if (nums==null || nums.length==0) return res;

11         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

12         map.put(0, -1);

13         int sum = 0;

14         for (int i=0; i<nums.length; i++) {

15             sum += nums[i];

16             if (!map.containsKey(sum)) {

17                 map.put(sum, i);

18             }

19             else {

20                 res.add(map.get(sum)+1);

21                 res.add(i);

22                 return res;

23             }

24         }

25         return res;

26     }

27 }

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

 1 class Element implements Comparable<Element>{

 2     int index;

 3     int value;

 4     public Element(int i, int v){

 5         index = i;

 6         value = v;

 7     }

 8     public int compareTo(Element other){

 9         return this.value-other.value;

10     }

11     public int getIndex(){

12         return index;

13     }

14     public int getValue(){

15         return value;

16     }

17 }

18 

19 public class Solution {

20     /**

21      * @param nums: A list of integers

22      * @return: A list of integers includes the index of the first number

23      *          and the index of the last number

24      */

25     public ArrayList<Integer> subarraySum(int[] nums) {

26         ArrayList<Integer> res = new ArrayList<Integer>();

27         if (nums==null || nums.length==0) return res;

28         int len = nums.length;

29         Element[] sums = new Element[len+1];

30         sums[0] = new Element(-1,0);

31         int sum = 0;

32         for (int i=0;i<len;i++){

33             sum += nums[i];

34             sums[i+1] = new Element(i,sum);

35         }

36         Arrays.sort(sums);

37         for (int i=0;i<len;i++)

38             if (sums[i].getValue()==sums[i+1].getValue()){

39                 int start = Math.min(sums[i].getIndex(),sums[i+1].getIndex())+1;

40                 int end = Math.max(sums[i].getIndex(),sums[i+1].getIndex());

41                 res.add(start);

42                 res.add(end);

43                 return res;

44             }

45 

46         return res;

47     }

48 }