[LeetCode] 560. Subarray Sum Equals K 子数组和为K

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:

1. The length of the array is in range [1, 20,000].
2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

解法1: Brute force. 用两丛循环i, j，计算sum[i, j]，如果等于K则记录结果。T: O(n^2)  S: O(1),  会TLE

解法2：Hashtable，基于公式 sum(i - j) = sum(0, j) - sum(0, i), 每循环一次都把数字累加到sum, 并用一个哈希表记录，key是sum， value是出现过的次数。当满足sum - K在哈希表中时，说明去掉之前那段和的数后剩下的数字和等于K, 满足条件，把记录的次数累加到结果(因为有多种组合可能)。T: O(n), S: O(n).

参考: https://discuss.leetcode.com/topic/87850/java-solution-presum-hashmap

Java:

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public class _560 {     public int subarraySum(int[] nums, int k) {         Map<Integer, Integer> preSum = new HashMap();         int sum = 0;         int result = 0;         preSum.put(0, 1);         for (int i = 0; i < nums.length; i++) {             sum += nums[i];             if (preSum.containsKey(sum - k)) {                 result += preSum.get(sum - k);             }             preSum.put(sum, preSum.getOrDefault(sum, 0) + 1);         }         return result;     }   }

Python:

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class Solution(object):     def subarraySum(self, nums, k):         """         :type nums: List[int]         :type k: int         :rtype: int         """         ans = sums = 0         cnt = collections.Counter()         for num in nums:             cnt[sums] += 1             sums += num             ans += cnt[sums - k]         return ans

Python:

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def subarraySum(self, nums, k):         count, cur, res = {0: 1}, 0, 0         for v in nums:             cur += v             res += count.get(cur - k, 0)             count[cur] = count.get(cur, 0) + 1         return res

Python: wo

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class Solution(object):     def subarraySum(self, nums, k):         """         :type nums: List[int]         :type k: int         :rtype: int         """         return res                res, sums = 0, 0         lookup = collections.Counter()         lookup[0] = 1 # important         for num in nums:             sums += num             if lookup[sums - k] > 0:                 res += lookup[sums - k]             lookup[sums] += 1                       return res

Python: TLE

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class Solution(object):     def subarraySum(self, nums, k):         """         :type nums: List[int]         :type k: int         :rtype: int         """         res = 0         for i in xrange(len(nums)):             sum = 0             for j in xrange(i, len(nums)):                 sum += nums[j]                   if sum == k:                     res += 1                           return res

C++:

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class Solution { public:     int subarraySum(vector<int>& nums, int k) {         int res = 0, n = nums.size();         for (int i = 0; i < n; ++i) {             int sum = nums[i];             if (sum == k) ++res;             for (int j = i + 1; j < n; ++j) {                 sum += nums[j];                 if (sum == k) ++res;             }         }         return res;     } };

C++:

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class Solution { public:     int subarraySum(vector<int>& nums, int k) {         int res = 0, sum = 0, n = nums.size();         unordered_map<int, int> m{{0, 1}};         for (int i = 0; i < n; ++i) {             sum += nums[i];             res += m[sum - k];             ++m[sum];         }         return res;     } };

　　　　

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