leetcode 81 Search in Rotated Sorted Array II ----- java

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.

第33题的延伸题。

33题说的是给定一个数组，这个数组可能从某个位置分成两部分，这两部分都是排序好的，假设没有重复的数字，给定一个数，然后求出那么数的位置，不存在则返回-1；

这道题说的就是假设可以重复。

当然不能直接循环一次，那样就没有意义了。

如果没有重复的数字，就修改一下二分法就行了，其实也就是几种情况，分类讨论一下就好。

在这里有重复的数字，那么就是，如果遇到pos和left处的数字一样，或者说pos和right处的数字一样，那么移动一个数字。

public class Solution {public boolean search(int[] nums, int target) {int len = nums.length;int left = 0,right = len-1;int pos;while( left <= right ){pos = (left+right)/2;if( target == nums[pos])return true;else if( nums[pos] > nums[left] ){if( target >= nums[left] && target < nums[pos])right = pos-1;elseleft = pos+1;}else if( nums[pos] < nums[right] ){if( target <= nums[right] && target > nums[pos])left = pos+1;elseright = pos-1;}else if( nums[pos] == nums[left] ){left++;}else if( nums[pos] == nums[right] )right--;}return false;}
}

转载于:https://www.cnblogs.com/xiaoba1203/p/5967285.html

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