Leetcode 744. Find Smallest Letter Greater Than Target

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本文详细解析了LeetCode 744题“查找大于目标字母的最小字符”的两种解题思路：线性搜索和二分搜索，并提供了C++代码实现。线性搜索时间复杂度为O(n)，而二分搜索则优化至O(logn)。

Leetcode 744. Find Smallest Letter Greater Than Target

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 = “j”
Output: “c”

题目大意：
给定一个只含有小写字母的数组和目标字母target，返回数组内最小的且大于目标字母target的元素。可以将26个字母视为一个环，比z大的字母为a。

解题思路1：
因为数组已经排好序，则只要遍历至letters[i] > target，该元素即为所求。若无，则说明target最大，返回第一个元素。时间复杂度O(n).

代码：

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        for(int i = 0; i < letters.size(); i++)
        {
            if(letters[i] > target)
                return letters[i];
        }
        return letters[0];
    }
};

解题思路2：
因为数组已排好序，再加上思路1之找到第一个比target大的位置，则可以使用二分搜索。时间复杂度可降为O(logn).

代码1 （比2快）：

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        int l = 0, r = letters.size() - 1;
        while(l <= r)
        {
            int mid = (r - l) / 2 + l; 
            if(letters[mid] == target)
            {
                if (mid + 1 != letters.size() && letters[mid + 1] != target)
                    return letters[mid + 1];
                else if(mid + 1 == letters.size())
                    return letters[0];
                else
                    l = mid + 1;
            }
            else if(letters[mid] < target)
                l = mid + 1;
            else 
                r = mid - 1;
        }
        return l == letters.size() ? letters[0] : letters[l];
    }
};

代码2 （精简）：

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        int l = 0, r = letters.size();
        while(l < r)
        {
            int mid = (r - l) / 2 + l; 
            if(letters[mid] <= target)
                l = mid + 1;
            else
                r = mid;
        }
        return letters[l % letters.size()];
    }
};