[抄题]:
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.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
用26数组,先加后减
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
有两种方法取出字母:.toCharArray(无参) 字符串.charAt()。
异或是个好东西。
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
//for loop, +t, -s
for (int i = 0; i < t.length(); i++) {
alpha[t.charAt(i) - 'a']++;
}
[其他解法]:
异或运算符是用符号“^”表示的,其运算规律是:
两个操作数的位中,相同则结果为0,不同则结果为1。
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
class Solution {
public char findTheDifference(String s, String t) {
//ini char[]
int[] alpha = new int[26]; //for loop, +t, -s
for (int i = 0; i < t.length(); i++) {
alpha[t.charAt(i) - 'a']++;
} for (int i = 0; i < s.length(); i++) {
alpha[s.charAt(i) - 'a']--;
} //return res
for (int j = 0; j < 26; j++) {
if (alpha[t.charAt(j) - 'a'] != 0) {
return t.charAt(j);
}
} return 0;
}
}