算法分析与设计week11--744. Find Smallest Letter Greater Than Target

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本文介绍了一种算法问题的解决方法，即在有序字符列表中找到最小的大于目标字母的元素。提供了两种解决方案，一种是简单的遍历比较方法，另一种是更高效的二分搜索方法。

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744. Find Smallest Letter Greater Than Target

Description

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’.

Example

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.

Analysis

解法1：暴力求解

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];
    }
};

解法2：二分搜索

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        if (target < letters[0] || letters[letters.size() - 1] <= target) return letters[0];

        int left = 0, right = letters.size() - 1, mid;

        while(left < right) {
            mid = (left + right) >> 1;
            if(target < letters[mid]) {
                right = mid;
            } else if (letters[mid] <= target) {
                left = mid + 1;
            }
        } 

        if (letters[mid] <= target) {
            return letters[mid + 1];
        } else if (letters[mid - 1] > target) {
            return letters[mid - 1];
        } else {
            return letters[mid];
        }
    }
};