问题
给定一个只包含小写字母的有序数组letters 和一个目标字母 target，寻找有序数组里面比目标字母大的最小字母。
数组里字母的顺序是循环的。举个例子，如果目标字母target = 'z' 并且有序数组为 letters = ['a', 'b']，则答案返回 'a'。
输入:
letters = ["c", "f", "j"]
target = "a"输出: "c"
输入:
letters = ["c", "f", "j"]
target = "c"输出: "f"
输入:
letters = ["c", "f", "j"]
target = "d"输出: "f"
输入:
letters = ["c", "f", "j"]
target = "g"输出: "j"
输入:
letters = ["c", "f", "j"]
target = "j"输出: "c"
输入:
letters = ["c", "f", "j"]
target = "k"输出: "c"
注:
- letters长度范围在[2, 10000]区间内。
- letters 仅由小写字母组成，最少包含两个不同的字母。
- 目标字母target 是一个小写字母。
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'.
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.
示例
public class Program {
public static void Main(string[] args) {
var letters = new char[] { 'c', 'f', 'j' };
var target = 'd';
var res = NextGreatestLetter(letters, target);
Console.WriteLine(res);
letters = new char[] { 'a', 'c', 'f', 'j' };
target = 'b';
res = NextGreatestLetter2(letters, target);
Console.WriteLine(res);
Console.ReadKey();
}
private static char NextGreatestLetter(char[] letters, char target) {
//暴力法
foreach(var c in letters) {
if(c > target) return c;
}
return letters[0];
}
private static char NextGreatestLetter2(char[] letters, char target) {
//二分法
var low = 0;
var high = letters.Length - 1;
var mid = 0;
while(low <= high) {
mid = low + (high - low) / 2;
if(letters[mid] > target) {
high = mid - 1;
} else if(letters[mid] <= target) {
low = mid + 1;
}
}
if(low >= letters.Length) low = 0;
return letters[low];
}
}以上给出2种算法实现，以下是这个案例的输出结果：
f
c分析：
显而易见，NextGreatestLetter 的时间复杂度为： ，NextGreatestLetter2 的时间复杂度为：
。