题目:原题链接(简单)
解法时间复杂度空间复杂度执行用时Ans 1 (Python) O ( N ) O(N) O(N) O ( 1 ) O(1) O(1)36ms (90.10%)Ans 2 (Python) O ( N ) O(N) O(N) O ( 1 ) O(1) O(1)52ms (14.87%)Ans 3 (Python)LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一:
def numJewelsInStones(self, J: str, S: str) -> int: ans = 0 for j in J: ans += S.count(j) return ans解法二:
def numJewelsInStones(self, J: str, S: str) -> int: ans = 0 for s in S: if s in J: ans += 1 return ans