leetcode 744. Find Smallest Letter Greater Than Target

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:

1. letters has a length in range [2, 10000].
2. letters consists of lowercase letters, and contains at least 2 unique letters.
3. target is a lowercase letter.

   class Solution(object):
       def nextGreatestLetter(self, letters, target):
           """
           :type letters: List[str]
           :type target: str
           :rtype: str
           """
           '''        
           Input:
   letters = ["c", "f", "j"]
   target = "j"
   Output: "c"
   
   Input:
   letters = ["c", "f", "j"]
   target = "k"
   Output: "c"
           just check the last letter if letters[-1] <= target: return letters[0]
           
           
           Input:
   letters = ["c", "f", "j"]
   target = "a"
   Output: "c"
           just check the first letter ....
   
   Input:
   letters = ["c", "f", "j"]
   target = "c"
   Output: "f"
   use binsearch mid letter > target and mid-letter <= target
   
   Input:
   letters = ["c", "f", "j"]
   target = "d"
   Output: "f"
   
   Input:
   letters = ["c", "f", "j"]
   target = "g"
   Output: "j"
           '''
           if letters[0] > target or letters[-1] <= target:
               return letters[0]
           
           i = 0
           j = len(letters)-1
           while i <= j:
               mid = (i+j) >> 1
               if letters[mid] <= target:
                   i = mid + 1
               else:
                   if letters[mid-1] <= target:
                       return letters[mid]
                   else:
                       j = mid -1                                        

   就是二分变种，没啥说的，但是要防止mid-1  < 0，因此，先判断了一下，确保target在letters里面。

4. ?

   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

   class Solution(object):     def nextGreatestLetter(self, letters, target):         """         :type letters: List[str]         :type target: str         :rtype: str         """         length = len(letters)           if letters[0] > target or letters[-1] <= target:             return letters[0]         i, j = 0, length-1         while i <= j:             mid = (i+j) >> 1             if letters[mid] <= target:                 i = mid + 1             else:                 j = mid - 1                                        return letters[i]

   本质，循环结束一定有，i=j+1, 在循环体里结束前一刻必然有，j==i==mid，走到判断那里，要么j=mid-1（满足target <letters[mid]），或者i=mid+1（letters[mid] <= target），所以一定有letters[j] <= target < letters[i]

5. 补充：bisect_right，

   1. #在L中查找x，x存在时返回x右侧的位置，
   2. x不存在返回应该插入的位置..

   
   

6. >>> import bisect
   >>> L = [1,3,3,6,8,12,15]
   >>> bisect.bisect_right(L,3)
   3
   >>> bisect.bisect_right(L,7)
   4
   >>> bisect.bisect_right(L,15)
   7
   >>> bisect.bisect_right(L,19)
   7
   >>> bisect.bisect_right(L,1)
   1
   >>> bisect.bisect_right(L,0)
   0