LWC 62：744. Find Smallest Letter Greater Than Target

LWC 62：744. Find Smallest Letter Greater Than Target

传送门：744. Find Smallest Letter Greater Than Target

Problem:

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.

思路：
因为wrap around，所以当给的target大于最大元素，就取集合中最小字符。否则，取大于target的第一个元素。（集合需要排序）

Java版：

    public char nextGreatestLetter(char[] letters, char target) {        Set<Character> set = new HashSet<>();        for (char c : letters) set.add(c);        char[] let = new char[set.size()];        int i = 0;        for (char c : set) {            let[i++] = c;        }        Arrays.sort(let);        for (int j = 0; j < let.length; ++j) {            if (let[j] > target) return let[j];        }        return let[0];    }

实际上是不需要去重的，代码如下：

    public char nextGreatestLetter(char[] letters, char target) {        Arrays.sort(letters);        for (int j = 0; j < letters.length; ++j) {            if (letters[j] > target) return letters[j];        }        return letters[0];    }

因为集合有序，所以可以使用upper_bound的二分查找，代码如下：

    public char nextGreatestLetter(char[] letters, char target) {        Arrays.sort(letters);        return letters[upperBound(letters, target)];    }    public int upperBound(char[] letters, char target) {        int l = 0;        int r = letters.length - 1;        while (l < r) {            int m = l + (r - l) / 2;            if (letters[m] <= target) {                l = m + 1;            }            else{                r = m;            }        }        if (letters[r] > target) return r;        else return 0;    }

当然你也可以使用Java自家的二分查找接口，但可惜的是，官方接口不支持重复元素集合的准确查找（会出错），所以在使用之前需要对元素去重。

代码如下：

    public char nextGreatestLetter(char[] letters, char target) {        Set<Character> set = new HashSet<>();        for (char c : letters) set.add(c);        char[] let = new char[set.size()];        int i = 0;        for (char c : set) {            let[i++] = c;        }        Arrays.sort(let);        int idx = Arrays.binarySearch(let, target);        if (idx >= 0){            if (idx + 1 >= let.length) return let[0];            else return let[idx + 1];        }        else {            idx = -idx - 1;            if (idx == let.length) return let[0];            else return let[idx];        }    }

Python版本：

    def nextGreatestLetter(self, letters, target):        """        :type letters: List[str]        :type target: str        :rtype: str        """        sorted(letters)        for l in letters:            if (target < l): return l        return letters[0]

自带的库也是非常方便：

    def nextGreatestLetter(self, letters, target):        """        :type letters: List[str]        :type target: str        :rtype: str        """        pos = bisect.bisect_right(letters, target)        return letters[0] if pos == len(letters) else letters[pos]