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For this problem we need to learn a new trick that if your start sum up all elements in an array. When you start from a to b to sum up all elements, if the result is zero, and also the summation of all elements from index a to c ( c<=b ) is zero. Then, we need to think about sub array from index c+1 to index b is a sub array with summation of all elements to zero.
Notice: we need to formalize the summation of index =-1 to 0. Then if there is a sub array start from 0 has summation of 0, we will return the result.
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> result = new ArrayList<Integer>(); 10 HashMap<Integer, Integer> map = new HashMap<Integer,Integer>(); 11 //set index -1 sum to be 0 then when you meet an index with sum equals 12 //0 you can return otherwise there is never a 0 stored for the empty 13 //list 14 map.put(0,-1); 15 int sum = 0; 16 for (int i = 0; i < nums.length; i++ ) { 17 sum += nums[i]; 18 if (map.containsKey(sum)) { 19 result.add(map.get(sum) + 1); 20 result.add(i); 21 return result; 22 } 23 map.put(sum,i); 24 } 25 return result; 26 } 27 }
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原文地址：http://www.cnblogs.com/ly91417/p/5789494.html
