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:
letters has a length in range [2, 10000].
letters consists of lowercase letters, and contains at least 2 unique letters.
target is a lowercase letter.
分析：用upper_bound返回第一个大于target的元素所在位置，如果这个位置等于letters.end()说明不存在，则返回letters的第一个值，否则返回it所在位置的元素值即可～
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class Solution { public: char nextGreatestLetter(vector<char>& letters, char target) { auto it = upper_bound(letters.begin(), letters.end(), target); return it == letters.end() ? letters[0] : *it; } }; |
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