Question

771. Jewels and Stones

中文文档

Description

You're given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels.

Letters are case sensitive, so "a" is considered a different type of stone from "A".

 

Example 1:

Input: jewels = "aA", stones = "aAAbbbb"
Output: 3

Example 2:

Input: jewels = "z", stones = "ZZ"
Output: 0

 

Constraints:

• 1 <= jewels.length, stones.length <= 50
• jewels and stones consist of only English letters.
• All the characters of jewels are unique.

Solutions

Solution 1

class Solution:
  def numJewelsInStones(self, jewels: str, stones: str) -> int:
    s = set(jewels)
    return sum(c in s for c in stones)

class Solution {
  public int numJewelsInStones(String jewels, String stones) {
    int[] s = new int[128];
    for (char c : jewels.toCharArray()) {
      s[c] = 1;
    }
    int ans = 0;
    for (char c : stones.toCharArray()) {
      ans += s[c];
    }
    return ans;
  }
}

class Solution {
public:
  int numJewelsInStones(string jewels, string stones) {
    int s[128] = {0};
    for (char c : jewels) s[c] = 1;
    int ans = 0;
    for (char c : stones) ans += s[c];
    return ans;
  }
};

func numJewelsInStones(jewels string, stones string) (ans int) {
  s := [128]int{}
  for _, c := range jewels {
    s[c] = 1
  }
  for _, c := range stones {
    ans += s[c]
  }
  return
}

function numJewelsInStones(jewels: string, stones: string): number {
  const s = new Set([...jewels]);
  let ans = 0;
  for (const c of stones) {
    s.has(c) && ans++;
  }
  return ans;
}

use std::collections::HashSet;
impl Solution {
  pub fn num_jewels_in_stones(jewels: String, stones: String) -> i32 {
    let mut set = jewels.as_bytes().iter().collect::<HashSet<&u8>>();
    let mut ans = 0;
    for c in stones.as_bytes() {
      if set.contains(c) {
        ans += 1;
      }
    }
    ans
  }
}

/**
 * @param {string} jewels
 * @param {string} stones
 * @return {number}
 */
var numJewelsInStones = function (jewels, stones) {
  const s = new Set(jewels.split(''));
  return stones.split('').reduce((prev, val) => prev + s.has(val), 0);
};

int numJewelsInStones(char* jewels, char* stones) {
  int set[128] = {0};
  for (int i = 0; jewels[i]; i++) {
    set[jewels[i]] = 1;
  }
  int ans = 0;
  for (int i = 0; stones[i]; i++) {
    set[stones[i]] && ans++;
  }
  return ans;
}