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].
Note
There is at least one subarray that it’s sum equals to zero.
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)
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
int len = nums.length;
ArrayList<Integer> ans = new ArrayList<Integer>();
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(0, -1);
int sum = 0;
for (int i = 0; i < len; i++) {
sum += nums[i];
if (map.containsKey(sum)) {
ans.add(map.get(sum) + 1);
ans.add(i);
return ans;
}
map.put(sum, i);
}
return ans;
}
}