LC.P560[和为K的子数组]

方法一：前缀和

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class Solution {
    public int subarraySum(int[] nums, int k) {
        int n = nums.length, ans = 0;
        int[] sum = new int[n + 1];
        for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + nums[i - 1];
        for (int i = 0; i < n; ++i) {
            for (int j = i; j < n; ++j) {
                if (sum[j + 1] - sum[i] == k) ++ans;
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(n^2)$
  • 空间复杂度：$O(n)$

方法二：前缀和+哈希表

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class Solution {
    public int subarraySum(int[] nums, int k) {
        int pre = 0, ans = 0;
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);
        for (int num : nums) {
            pre += num;
            if (map.containsKey(pre - k)) {
                ans += map.get(pre - k);
            }
            map.merge(pre, 1, Integer::sum);
        }
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$