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Formatted question description: https://leetcode.ca/all/744.html

# 744. Find Smallest Letter Greater Than Target (Easy)

Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.

Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.

Examples:

Input:
letters = ["c", "f", "j"]
target = "a"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "c"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "d"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "g"
Output: "j"

Input:
letters = ["c", "f", "j"]
target = "j"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "k"
Output: "c"


Note:

1. letters has a length in range [2, 10000].
2. letters consists of lowercase letters, and contains at least 2 unique letters.
3. target is a lowercase letter.

## Solution 1.

• class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int length = letters.length;
char lastLetter = letters[length - 1];
if (lastLetter <= target)
return letters[0];
int index = binarySearch(letters, target);
return letters[index];
}

public int binarySearch(char[] letters, char target) {
target++;
int low = 0, high = letters.length - 1;
while (low <= high) {
int mid = (high - low) / 2 + low;
char c = letters[mid];
if (c == target)
return mid;
else if (c > target)
high = mid - 1;
else
low = mid + 1;
}
return low;
}
}

############

class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int left = 0, right = letters.length;
while (left < right) {
int mid = (left + right) >> 1;
if (letters[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return letters[left % letters.length];
}
}

• // OJ: https://leetcode.com/problems/find-smallest-letter-greater-than-target/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
char nextGreatestLetter(vector<char>& letters, char target) {
auto i = upper_bound(letters.begin(), letters.end(), target);
return i == letters.end() ? letters[0] : *i;
}
};

• class Solution:
def nextGreatestLetter(self, letters: List[str], target: str) -> str:
left, right = 0, len(letters)
while left < right:
mid = (left + right) >> 1
if ord(letters[mid]) > ord(target):
right = mid
else:
left = mid + 1
return letters[left % len(letters)]

############

class Solution(object):
def nextGreatestLetter(self, letters, target):
"""
:type letters: List[str]
:type target: str
:rtype: str
"""
for letter in letters:
## 提交了之后发现不用使用ord，字符可以用'>''<'比较大小
if ord(letter) > ord(target):
return letter
return letters[0]

• func nextGreatestLetter(letters []byte, target byte) byte {
left, right := 0, len(letters)
for left < right {
mid := (left + right) >> 1
if letters[mid] > target {
right = mid
} else {
left = mid + 1
}
}
return letters[left%len(letters)]
}

• function nextGreatestLetter(letters: string[], target: string): string {
const n = letters.length;
let left = 0;
let right = letters.length;
while (left < right) {
let mid = (left + right) >>> 1;
if (letters[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return letters[left % n];
}


• impl Solution {
pub fn next_greatest_letter(letters: Vec<char>, target: char) -> char {
let n = letters.len();
let mut left = 0;
let mut right = n;
while left < right {
let mid = left + (right - left) / 2;
if letters[mid] > target {
right = mid;
} else {
left = mid + 1;
}
}
letters[left % n]
}
}


• class Solution {
/**
* @param String[] $letters * @param String$target
* @return String
*/
function nextGreatestLetter($letters,$target) {
$left = 0;$right = count($letters); while ($left <= $right) {$mid = floor($left + ($right - $left) / 2); if ($letters[$mid] >$target) $right =$mid - 1;
else $left =$mid + 1;
}
if ($left >= count($letters)) return $letters[0]; else return$letters[\$left];
}
}