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Search in Rotated Sorted Array II

时间:2014-08-23 02:15:59      阅读:157      评论:0      收藏:0      [点我收藏+]

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Follow up for "Search in Rotated Sorted Array":

What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

思路:与“Search in Rotated Sorted Array”类似。只需增加考虑A[start]==A[end]的情况即可,因为此时存在不确定的情况:A[middle]==A[start],此时,我们无法确定A[middle]是来自第一个序列还是第二个序列。

 1 class Solution {
 2 public:
 3     bool search( int A[], int n, int target ) {
 4         int start = 0, end = n-1;
 5         while( start <= end ) {
 6             int middle = ( start + end ) / 2;
 7             if( A[middle] == target || A[start] == target || A[end] == target ) { return true; }
 8             if( A[start] < A[end] ) {
 9                 if( A[middle] < target ) {
10                     start = middle + 1;
11                     --end;
12                 } else {
13                     end = middle - 1;
14                     ++start;
15                 }
16             } else if( A[start] > A[end] ){
17                 if( A[middle] < target ) {
18                     if( target > A[start] && A[middle] < A[start] ) {
19                         end = middle - 1;
20                         ++start;
21                     } else {
22                         start = middle + 1;
23                         --end;
24                     }
25                 } else {
26                     if( target < A[start] && A[middle] >= A[start] ) {
27                         start = middle + 1;
28                         --end;
29                     } else {
30                         end = middle - 1;
31                         ++start;
32                     }
33                 }
34             } else {
35                 ++start; --end;
36             }
37         }
38         return false;
39     }
40 };

 

Search in Rotated Sorted Array II

标签:style   blog   color   io   for   ar   art   div   cti   

原文地址:http://www.cnblogs.com/moderate-fish/p/3930287.html

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