Search in Rotated Sorted Array II
Search in Rotated Sorted Array II
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.
这道题说如果有重复的元素会不会对，时间之类的有影响，对于我这种懒人直接copy上一题的过去A了，不过感觉和线性搜索O（n）的时间复杂度提升不是太多，只是后面的用了二分查找O（logn）
虽然是一样的，还是贴一下代码，恩，我就是这么无聊
1 public class Solution { 2 public boolean search(int[] A, int target) { 3 int i = 0; 4 for(; i < A.length; i++){ 5 if(i > 0 && A[i - 1] > A[i]) //找出轴的位置 6 break; 7 if(A[i] == target) 8 return true; 9 } 10 if(i == A.length) //没有找到 11 return false; 12 //从轴的位置到最后开始二分查找从i的位置开始到length - 1 13 int low = i; 14 int high = A.length - 1; 15 int middle; 16 while(low <= high){ 17 middle = (low + high) / 2; 18 if(A[middle] == target) 19 return true; 20 else if(A[middle] > target){ //前半段找 21 high = middle - 1; 22 }else{ //后半段找 23 low = middle + 1; 24 } 25 } 26 27 return false; 28 } 29 }