Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
Example:
Input: s = "abcd" t = "abcde" Output: e Explanation: 'e' is the letter that was added.
class Solution { public char findTheDifference(String s, String t) { List<Character> list = new ArrayList(); for(char c: t.toCharArray()){ list.add(c); } for(char c: s.toCharArray()){ if(list.indexOf(c) >= 0){ list.remove(list.indexOf(c)); } } return list.get(0); } }
我怎么老把遍历对象搞错卧槽?
想法是:把t存起来,然后遍历s,发现当前char在t中就remove,最后剩下的就是答案
class Solution { public char findTheDifference(String s, String t) { int ss = 0; int tt = 0; for(char c: s.toCharArray()) ss += (int) c; for(char c: t.toCharArray()) tt += (int) c; return (char) (tt - ss); } }
刚还在那想能不能两个一减,卧槽真能