You're given strings?J?representing the types of stones that are jewels, and?S?representing the stones you have.? Each character in?Sis a type of stone you have.? You want to know how many of the stones you have are also jewels.
The letters in?J?are guaranteed distinct, and all characters in?J?and?S?are letters. Letters are case sensitive, so?"a"?is considered a different type of stone from?"A".
Example 1:
Input:J = "aA", S = "aAAbbbb"Output:3
Example 2:
Input:J = "z", S = "ZZ"Output:0
Note:
S?and?J?will consist of letters and have length at most 50.
The characters in?J?are distinct.
Solution：（錯(cuò)誤解法）
https://discuss.leetcode.com/topic/118697/java-o-s-j-time-and-o-1-space
public int numJewelsInStones(String J, String S) {
? ? ? ? if(J.length() == 0 || S.length() == 0)
? ? ? ? ? ? return 0;
? ? ? ? // Using helper array for O(1) loopkup when traversing S
? ? ? ? int[] jewels = new int[58];
? ? ? ? for(int i=0; i
? ? ? ? ? ? jewels[(J.charAt(i) - 'A')] = 1;
? ? ? ? }
? ? ? ? int result = 0;
? ? ? ? for(int i=0; i
? ? ? ? ? ? if(jewels[(S.charAt(i) - 'A')] == 1) {
? ? ? ? ? ? ? ? result++;
? ? ? ? ? ? }
? ? ? ? }
? ? ? ? return result;
? ? }
}
Why an array of length 58? Two reasons
We need to consider both upper-case and lower-case characters
ASCII of 'z' - 'A' is 58. (https://www.cs.cmu.edu/~pattis/15-1XX/common/handouts/ascii.html)
Thanks to?@thaliahard?for pointing out the reduction in array size.
