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  • [LeetCode] 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 = "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.

    找出比目标值大的最小的那个字母

    给定一个字符数组,找出比目标值大的最小的那个字符,如果目标值大于数组中最大的字母,则返回数组中第一个字母。

    使用upper_bound()这个函数找出比目标字母大的那个数,这时需要判断

    如果目标值大于等于数组中最大的字母,则返回数组中的首字母。

    否则返回找到的那个字母。

    class Solution {
    public:
        char nextGreatestLetter(vector<char>& letters, char target) {
            auto it = upper_bound(letters.begin(), letters.end(), target);
            if (it == letters.end())
                return letters.front();
            else {
                return *it;
            }
        }
    };
    // 12 ms

    也可以使用二分搜索。

    class Solution {
    public:
        char nextGreatestLetter(vector<char>& letters, char target) {
            if (letters.back() <= target)
                return letters.front();
            int left = 0, right = letters.size() - 1;
            while (left < right)  {
                int mid = left + (right - left) / 2;
                if (letters[mid] > target) {
                    right = mid;
                }
                else {
                    left = mid + 1;
                }
            }
            return letters[left];
        }
    };
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  • 原文地址:https://www.cnblogs.com/immjc/p/8027263.html
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