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33. Search in Rotated Sorted Array (Array;Divide-and-Conquer)
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.
思路: 改动的二分法。有三种情况：正序、右侧rotate、左侧rotate。三种情况分别讨论。
class Solution {public: int search(vector<int>& nums, int target) { return binarySearch(nums,0,nums.size()-1, target); } int binarySearch(vector<int>& nums, int start, int end, int target){ if(start>end){ return -1; } int mid = start+ ((end-start)>>1); if(nums[mid]==target) return mid; //正序 if(nums[mid]>=nums[start] && nums[mid]<nums[end]){ //mid可能会=start，所以这里要用>= if(target < nums[mid]) return binarySearch(nums,start,mid-1,target); //mid+1或-1,每次至少舍弃一个数 else return binarySearch(nums,mid+1,end,target); } //右侧rotate else if(nums[mid]>=nums[start] && nums[mid]>nums[end]){ if(target>=nums[start] && target<nums[mid]) return binarySearch(nums,start,mid-1,target); else return binarySearch(nums,mid+1,end,target); } //左侧rotate else{ if(target>=nums[start] || target<nums[mid]) return binarySearch(nums,start,mid-1,target); else return binarySearch(nums,mid+1,end,target); } }};
33. Search in Rotated Sorted Array (Array;Divide-and-Conquer)
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