744 Find Smallest Letter Greater Than Target 寻找比目标字母大的最小字母
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'.
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.
题目描述:
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'。
示例 :
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。
思路:
由于数组有序, 用二分查找, 注意顺序是循环的, 所以比较一下数组的最后一个元素和 target的大小来判定返回值
时间复杂度O(lgn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
char nextGreatestLetter(vector<char>& letters, char target)
{
int i = 0, j = letters.size() - 1;
while (i <= j)
{
int mid = i + ((j - i) >> 1);
if (letters[mid] > target) j = mid - 1;
else i = mid + 1;
}
return letters[letters.size() - 1] <= target ? letters[0] : letters[i];
}
};
Java:
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int i = 0, j = letters.length - 1;
while (i <= j) {
int mid = i + ((j - i) >> 1);
if (letters[mid] > target) j = mid - 1;
else i = mid + 1;
}
return letters[letters.length - 1] <= target ? letters[0] : letters[i];
}
}
Python:
class Solution:
def nextGreatestLetter(self, letters: List[str], target: str) -> str:
return letters[[x > target for x in letters].index(max([x > target for x in letters]))]
