[LeetCode] Search in Rotated Sorted Array II
分类:c++oj
c++leetcode
Search in Rotated Sorted Array II
Follow up for "Search in Rotated Sorted Array":What ifduplicatesare allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
解题思路:
同Search in Rotated Sorted Array一样,这道题也可以用二分查找。不同的是,这道题元素可以相同,因此,在最坏的情况下时间复杂度为O(n)。
在每次二分查找之前,先压缩左右区间。然后判断:
1、当nums[left]、nums[middle]、nums[right]之一与target相同时,返回true
2、当target<nums[middle]时,分middle在大部还是小部两种情况:
(1)若middle在大部,即nums[middle] > nums[left],则看nums[left]的大小。若nums[left]>target,那么left~middle可以排除,让left=middle+1。若nums[left]<target,那么target的值只能存在left~middle之间,让right=middle – 1
(2)若middle在小部,即nums[middle] < nums[left],那么middle~right可以排除,令right = middle – 1
3、同样,当target>nums[middle]时,,分middle在大部还是小部两种情况:
(1)若middle在大部,即nums[middle]>nums[left],那么left~middle可以排除,令left=middle + 1
(2)若middle在小部,即nums[middle]<nums[left]。如果nums[right]>target,那么target可定在middle~right之间,令left=middle+1。否则,nums[right]<target,那么可以排除middle~right之间的元素,令right = middle – 1
下面是代码:
class Solution {public:bool search(vector<int>& nums, int target) {int len = nums.size();int left = 0;int right = nums.size() – 1;while(left <= right){while(left < right && left<len – 1 && nums[left] == nums[left+1]){left++;}while(left < right && right>=1 && nums[right] == nums[right-1]){right–;}int middle = (left + right) / 2;if(target == nums[middle] || target == nums[left] || target == nums[right]){return true;}else if(target < nums[middle]){if(nums[middle]>nums[left]){ //表明middle在大部if(nums[left] > target){left = middle + 1;}else{right = middle – 1;}}else{//表明middle在小部right = middle – 1;}}else{if(nums[middle]>nums[left]){ //表明middle在大部left = middle + 1;}else{//表明middle在小部if(nums[right]>target){left = middle + 1;}else{right = middle – 1;}}}}return false;}};
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