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 1:
Input: [-3, 1, 2, -3, 4]
Output: [0, 2] or [1, 3].
Explanation: return anyone that the sum is 0.
Example 2:
Input: [-3, 1, -4, 2, -3, 4] Output: [1,5]
思路：某一段[l, r]的和为0， 则其对应presum[l-1] = presum[r].
presum 为数组前缀和。只要保存每个前缀和，找是否有相同的前缀和即可。
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;
}
}
原文：https://www.cnblogs.com/FLAGyuri/p/12078472.html