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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];
    }
}

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