lintcode-138-子数组之和

138-子数组之和

给定一个整数数组，找到和为零的子数组。你的代码应该返回满足要求的子数组的起始位置和结束位置

注意事项

There is at least one subarray that it's sum equals to zero.

样例

给出 [-3, 1, 2, -3, 4]，返回[0, 2] 或者 [1, 3].

标签

子数组 哈希表

方法一（O(n^2)）

利用两个for循环，每次取出一个元素依次与后面的元素相加，时间复杂度是O(n^2)

code

class Solution {
public:/*** @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*/vector<int> subarraySum(vector<int> nums){// write your code hereint size = nums.size(), sum = 0;if(size <= 0) {return vector<int>();}vector<int> result;for(int i=0; i<size; i++) {sum = nums[i];if(sum == 0){result.push_back(i);result.push_back(i);return result;}for(int j=i+1; j<size; j++) {sum += nums[j];if(sum == 0) {result.push_back(i);result.push_back(j);return result;}}}return result;}
};

方法二（O(n)）

利用一个 map 记录从第一个元素开始到当前元素之和 与 当前元素的下标 的对应关系，若有一段子数组和为0，那么势必出现同一和对应2个下标，此时就找到了和为零的连续序列，时间复杂度是O(n)

code

class Solution {
public:/*** @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*/vector<int> subarraySum(vector<int> nums){// write your code hereint size = nums.size(), sum = 0;if(size <= 0) {return vector<int>();}vector<int> result;map<int, int> subSum;subSum[0] = -1;for(int i=0; i<size; i++) {sum += nums[i];if(subSum.count(sum)) {result.push_back(subSum[sum] + 1);result.push_back(i);return result;}subSum[sum] = i;}return result;}
};