Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Example 1:
Input:nums = [1,1,1], k = 2 Output: 2
Note:
- The length of the array is in range [1, 20,000].
- The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
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笨方法就是遍历所有子序列,由于起点和终点都要遍历,所以时间复杂度是O(n^2)
本次的方法是遍历所有起点是最开始元素的子序列,我们起名叫sum[i],只要两个sum[i]的差值是k,则两个i作为起点和终点的子序列就符合要求。由于起点就是最开始元素,只需遍历终点,时间复杂度为O(n)。
public class Solution { public int subarraySum(int[] nums, int k) { Map<Integer,Integer> map = new HashMap<Integer,Integer>(); map.put(0,1); int sum = 0; int res = 0; for(int i = 0;i<nums.length;i++) { sum += nums[i]; res += map.getOrDefault(sum-k,0); map.put(sum,map.getOrDefault(sum,0)+1); } return res; } }