Solution 1:

直接累加。统计和

$2 3, 4 5 7 11 12 13 14 15 16 17 18 class Solution { //OCnA2\) public int subarraySum$intC\] nums, int ans = 0; int n — nums.length; for\(int i int sum = 0; for\(int j=i; j<n; sum numsCj\]; i fCsum ans 1; int k$ { //no break since array is not sort in order. return ans;$

Solution 2:Map

O(n)

存sum的次数

Sum2 - Sum1 = k

找sum1的次数

$O Java class Solution { 2, 3 4 5 6 10 11 12 13 v 14 15 16 v 17 18 19 20 21 22 public int subarraySum$intC\] nums, int k$ { int count Map<lnteger, Integer> sumCntMap = new HashMap<>; sumCntMap . put$Ø, 1$; int sum = 0; for$int i=0; i < nums. length; i++${ sum numsCi\] ; i f$sumCntMap. contai nsKey\(sum- k$\) { count sumCntMap.getCsum-k\); i f$sumCntMap. contai sumCntMap. put\(sum , else sumCntMap. put\(sum , return count; sumCntMap.get\(sum$ + 1\); 1\);$

Map

Solution 1:

Solution 2:Map

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