Subarray Sum

原题链接：

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.

样例

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

标签

SOLUTION 1:

我们有一个O(N)的解法。使用Map 来记录index, sum的值。当遇到两个index的sum相同时，表示从index1+1到index2是一个解。

注意：添加一个index = -1作为虚拟节点。这样我们才可以记录index1 = 0的解。

空间复杂度：O(N)

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
      
        subarraySum(int[] nums) { 8         // write your code here 9         10         int len = nums.length;11         12         ArrayList
       
         ret = new ArrayList
        
         ();13         14         HashMap
         
           map = new HashMap
          
           ();15         16         // We set the index -1 sum to be 0 to let us more convient to count.17         map.put(0, -1);18         19         int sum = 0;20         for (int i = 0; i < len; i++) {21             sum += nums[i];22             23             if (map.containsKey(sum)) {24                 // For example: 25                 //        -3  1  2 -3 426                 // SUM: 0 -3 -2  0 -3 127                 // then we got the solution is : 0 - 228                 ret.add(map.get(sum) + 1);29                 ret.add(i);30                 return ret;31             }32             33             // Store the key:value of sum:index.34             map.put(sum, i);35         }36         37         return ret;38     }39 }
          
         
        
       
      

 

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