[LeetCode]Search in Rotated Sorted Array II

Search in Rotated Sorted Array II

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Total Accepted: 50992 Total Submissions: 162671 Difficulty: Medium

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.

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这个题目我想了好久才过的，本身循环排序数据就不是很好做。这个由于有重复数，所以当重复数是最前面和最后面的时候需要进行递归。

public class Solution {    public boolean search(int[] nums, int target) {return search(nums, target, 0, nums.length-1);}public boolean search(int[] nums, int target,int m,int n) { if(m>n)return false; if(m<0||n>=nums.length)return false; int i=m;int j=n; if(nums[m]==target)return true;         while(i<=j){         int mid = (i+j)/2;         if(nums[mid]==target){         return true;         }         else{         if(nums[mid]<target){         if(nums[mid]>nums[m]){         i=mid+1;         }         else{         if(nums[mid]<nums[m]){         if(target<nums[m]){         i=mid+1;         }         else{        j=mid-1;         }         }         else{         if(search(nums, target, m+1, mid-1)||search(nums, target, mid+1, n)){         return true;         }         else{         return false;         }         }         }         }         else{         if(nums[mid]<nums[m]){         j=mid-1;         }         else{         if(nums[mid]>nums[m]){         if(target>nums[m]){         j=mid-1;         }         else{         i=mid+1;         }         }         else{         if(search(nums, target, m+1, mid-1)||search(nums, target, mid+1, n)){         return true;         }         else{         return false;         }         }         }         }         }         }         return false;    }}