每次二分有四种情况:
1. nums[mid] = target,则可以返回mid
2. nums[mid] < nums[right],说明在[mid, right]区间是右边递增的区间,然后判断target是否在这个区间内
1)如果nums[mid] < target <= nums[right],说明target在右边区间里,则left = mid + 1;
2)否则在左边区间里,搜索左边区间,right = mid - 1;
3. nums[mid] > nums[right],说明[elft, mid]区间是在左边的递增区间,然后判断target是否在这个左边区间里
1)如果nums[left] <= target < nums[mid],说明target在这个区间里,则使right = mid - 1;
2)否则说明target在[mid, right]的不规则区间里,搜索右边区间,则使left = mid + 1;
4. nums[mid] == nums[right],这种情况并不知道搜索哪一边,因为nums[right] != target,此时可以抛弃nums[right],搜索[left, right-1]的区域
class Solution { public: bool searchD(vector<int> & nums, int target, int s, int e) { cout<<s<<" "<<e<<endl; if(s>e||(s==e&& nums[s]!=target)) return false; int mid=s+(e-s)/2; if(target==nums[mid]) return true; // cout<<s<<" "<<e<<" "<<mid<<endl; // cout<<nums[mid]<<" "<<nums[s]<<" "<<nums[e]<<endl; if(nums[s]==target) return true; if(nums[e]==target) return true; if(nums[mid]==nums[s]&& nums[mid==nums[e]]) return searchD(nums, target, s+1,e-1); if(nums[mid]>nums[e]) { if(nums[mid]>target&&nums[s]<target ) { // cout<<'1'<<endl; return searchD(nums, target,s, mid); } else { // cout<<'2'<<endl; return searchD(nums,target, mid+1,e); } } else { if(nums[mid]<target&& target<nums[e]) { // cout<<'3'<<endl; return searchD(nums,target,mid+1, e); } else { // cout<<'4'<<endl; return searchD(nums,target,s, mid); } } } bool search(vector<int>& nums, int target) { int size=nums.size(); if(size<1) return false; bool result=searchD(nums, target, 0, size-1); return result; } };