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。
第一种思路:
线性扫描整个数组,如果遇到比target大的item就直接返回item,
如果过程中没有遇到比target大的就说明应该返回letters[0]
class Solution(object):
def nextGreatestLetter(self, letters, target):
"""
:type letters: List[str]
:type target: str
:rtype: str
"""
# lo, hi = 0, len(letters) - 1
for letter in letters:
if ord(letter) > ord(target):
return letter
return letters[0]
第二种思路:
二分查找,先把极端情况去掉然后查找。
class Solution(object):
def nextGreatestLetter(self, letters, target):
"""
:type letters: List[str]
:type target: str
:rtype: str
"""
lo, hi = 0, len(letters) - 1
if target < letters[0] or target >= letters[-1]:
return letters[0]
while(lo <= hi):
mid = lo + (hi - lo) // 2
if letters[mid] <= target:
lo = mid + 1
else:
hi = mid - 1
return letters[lo]