【LeetCode】Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Discuss

java code : 找到那个 pivot ，然后二分查找， 找pivot的算法复杂度也是 O（logn）：

public class Solution {    public int search(int[] A, int target) {        // Note: The Solution object is instantiated only once and is reused by each test case.       if(A == null)            return -1;        int lhs = 0, rhs = A.length - 1;        int pos = 1;        int mid = lhs;        if(A[lhs] <= A[rhs])        {        int index = Arrays.binarySearch(A, target);        if(index < 0)        return -1;        return index;        }        while(A[lhs] >= A[rhs])        {        if(rhs - lhs == 1)        {        pos = rhs;        break;        }        mid = (lhs + rhs) >> 1;        if(A[mid] >= A[lhs])        lhs = mid;        else if(A[mid] <= A[rhs])        rhs = mid;        }        int res1 = binarysearch(A,target,0,pos-1);        int res2 = binarysearch(A,target,pos,A.length -1);        if(res1 != -1)        return res1;        if(res2 != -1)        return res2;        return -1;    }    public int binarysearch(int[] array, int key, int st, int end){int lhs = st, rhs = end;if(st == end){if(array[st] == key)return st;return -1;}while(lhs <= rhs){int mid = (lhs + rhs) >> 1;if(array[mid] == key)return mid;else if(array[mid] < key)lhs = mid + 1;else rhs = mid - 1;}return -1;}}

Search in Rotated Sorted Array II

 Total Accepted: 1164 Total Submissions: 4138My Submissions

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.

有重复的情况，只能朴素查找了，会影响算法的复杂度，因为会影响找pivot的那个算法

java code :

public class Solution {    public boolean search(int[] A, int target) {        // Note: The Solution object is instantiated only once and is reused by each test case.        if(A == null)              return false;          int lhs = 0, rhs = A.length - 1;          int pos = 1;          int mid = lhs;          if(A[lhs] <= A[rhs])          {               Arrays.sort(A);            int index = Arrays.binarySearch(A, target);              if(index < 0)                  return false;              return true;          }          while(A[lhs] >= A[rhs])          {              if(rhs - lhs == 1)              {                  pos = rhs;                  break;              }              mid = (lhs + rhs) >> 1;              if(A[mid] == A[lhs] && A[lhs] == A[rhs])            {                Arrays.sort(A);                int index = Arrays.binarySearch(A, target);                  if(index < 0)                      return false;                  return true;              }            if(A[mid] >= A[lhs])                  lhs = mid;              else if(A[mid] <= A[rhs])                  rhs = mid;          }          int res1 = binarysearch(A,target,0,pos-1);          int res2 = binarysearch(A,target,pos,A.length -1);          if(res1 != -1)              return true;          if(res2 != -1)              return true;          return false;      }          public int binarysearch(int[] array, int key, int st, int end)      {          int lhs = st, rhs = end;          if(st == end)          {              if(array[st] == key)                  return st;              return -1;          }          while(lhs <= rhs)          {              int mid = (lhs + rhs) >> 1;              if(array[mid] == key)                  return mid;              else if(array[mid] < key)                  lhs = mid + 1;              else rhs = mid - 1;          }          return -1;      }    }