[leetcode] 560. Subarray Sum Equals K
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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 = 2Output: 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].
这道题是统计相加和为k的连续子数组个数,题目难度为Medium。
题目和第523题有些类似,感兴趣的同学可以顺便看下第523题(传送门)。
最直观的想法是遍历数组并依次加当前位置的数字,同时用数组sum记录下当前位置之前所有数字的相加和,这样下标[i, j)之间的数字之和就可以用sum[j]-sum[i]来计算,然后通过双层循环,遍历所有情况来统计满足条件的子数组个数。具体代码:
class Solution {public: int subarraySum(vector<int>& nums, int k) { int sz = nums.size(), cnt = 0; vector<int> sum(sz+1, 0); for(int i=0; i<sz; ++i) sum[i+1] = sum[i] + nums[i]; for(int i=0; i<sz; ++i) { for(int j=i+1; j<=sz; ++j) { if(sum[j] - sum[i] == k) ++cnt; } } return cnt; }};
用sum表示从数组开始位置到当前位置的数字相加和,有了第523题的经验,我们还可以用Hash Table来存储sum出现的次数,如果当前位置之前有相加和为sum-k的位置,则这两个位置之间的数字相加和为k,以当前位置结尾的相加和为k的子数组个数为hash[sum-k],这样遍历整个数组即可得出满足条件的子数组个数。具体代码:
class Solution {public: int subarraySum(vector<int>& nums, int k) { int sum = 0, cnt = 0; unordered_map<int, int> hash; hash[0] = 1; for(auto n:nums) { sum += n; cnt += hash[sum-k]; ++hash[sum]; } return cnt; }};
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