# Daily practice (47): Find the difference

title: Daily Practice (47): Find the Difference

categories:[Swords offer]

tags:[Daily practice]

date: 2022/04/22

## Daily practice (47): Find the difference

Given two strings s and t, they contain only lowercase letters. String t is randomly rearranged from string s, and then a letter is added at a random position. Find the letter that was added in the t.

Example 1:

Input: s = "abcd", t = "abcde"

output: "e"

Explanation: 'e' is the letter that was added.

Example 2:

Input: s = "", t = "y"

output: "y"

hint:

0 <= s.length <= 1000

t.length == s.length + 1

s and t contain only lowercase letters

Source: LeetCode

### Method 1: Summation

#### Thought analysis

This question is to add only one character, find the ASCII value of the s string and the ASCII value of t and then the ASCII value of t - s to get the ASCII value of the added character

``` ```char findTheDifference(string s, string t) {
int as = 0, at = 0;
for (auto ch : s) {
as += ch;
}
for (auto ch : t) {
at += ch;
}
return at - as;
}``````

### Method 2: Bit Operations

#### Thought analysis

If two strings are concatenated into one string, the problem turns into finding the odd number of characters in the string

Features of the XOR operation:

• XOR itself to 0, and XOR 0 to itself;
• It has commutative and associative laws, such as 1^2^3^4^2^3^1 = (1^1)^(2^2)^(3^3)^4 = 0^0^0^4 = 0^4 = 4;
• Summary: The XOR operation is good at finding differences.
``` ```char findTheDifference(string s, string t) {
int ret = 0;
for (auto ch : s) {
ret ^= ch;
}
for (auto ch : t) {
ret ^= ch;
}
return ret;
}``````

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