leetcode rotated sorted array

1. Find minimum in rotated sorted array(https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array/)

解题思路： L为array左端，R为array右端，M = (L+R)/2，若num[M] > num[R], 则说明rotated的部分在右边，最小值为M+1 ~ R中的一个元素。若num[M] < num[R], 则说明rotated的部分在左边，L ～ M中的一个元素可能为最小值。

 1   def findMin(self, num):
 2     lenNum = len(num)
 3     L = 0
 4     R = lenNum - 1
 5     while L < R and num[L] > num[R]:
 6       M = (L+R)/2
 7       if num[M] > num[R]:
 8         L = M + 1
 9       else:
10         R = M
11     return num[L]

2. Find Minimum in Rotated Sorted Array II(https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)

解题思路：此题与上一题的不同之处在于数组里有重复的数字。判定条件需要变一下，如果num[L] >= num[R]则存在rotated， =号的存在就是由于有重复的数字。若出现num[M] == num[R]或num[L] == num[M]的情况，均不能判定M～R或L～M之间是否存在rotated，因此L++, 只判定不等于的情况。

 1   def findMin(self, num):
 2     lenNum = len(num)
 3     L = 0
 4     R = lenNum - 1
 5     while L < R and num[L] >= num[R]:
 6       M = (L + R)/2
 7       if num[M] > num[R]:
 8         L = M + 1
 9       elif num[M] < num[L]:
10         R = M
11       else:
12         L += 1
13     return num[L]

3. Search In Rotated Sorted Array(https://oj.leetcode.com/problems/search-in-rotated-sorted-array/)

解题思路：二分法

 1   def search(self, A, target):
 2     lenN = len(A)
 3     l = 0
 4     r = lenN - 1
 5     while l <= r:
 6       m = (l+r)/2
 7       if A[m] == target:
 8         return m
 9       if A[l] <= target < A[m]:
10         r = m - 1
11       elif A[m] < A[r] and (target < A[m] or target > A[r]):
12         r = m - 1
13       else:
14         l = m + 1
15     return -1

4. Search In Rotated Sorted Array II(https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/)

解题思路：A[L] < A[M]才表示右边部分rotated，若A[L] == A[M]，则不能保证一定是右边部分rotated，此时L+= 1

 1   def search(self, A, target):
 2     lenN = len(A)
 3     l = 0
 4     r = lenN - 1
 5     while l <= r:
 6       m = (l+r)/2
 7       if A[m] == target:
 8         return True
 9       if A[l] < A[m]:
10         if A[l] <= target < A[m]:
11           r = m - 1
12         else:
13           l = m + 1
14       elif A[l] > A[m]:
15         if A[m] < target <= A[r]:
16           l = m + 1
17         else:
18           r = m - 1
19       else:
20         l += 1
21     return False

 

posted @ 2014-10-28 20:39  darlwen  阅读(125)  评论(0)  编辑  收藏  举报