*Leetcode 81. Search in Rotated Sorted Array II

常规容易想到的解法：

基本就是两点：1、判断在两个上升的段的哪个段 2、mid跟r相等的时候r--可以解决问题的

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        if(nums.size() == 0) return false;
        
        int l = 0, r = nums.size()-1;
        //int end = nums.size() - 1;
        while (l <= r) {
            int mid = (l + r)/2;
            
            if (target == nums[r] || target == nums[mid]) return true;
            
            
            if (nums[mid] == nums[r]) {
                // for (int j = l; j <= r; j ++) {
                //         if (nums[j] == target) return true;
                //     } 
                //     return false;
                r --;
                continue;
            }
            
            if (nums[mid] > nums[r]) {
                if (target < nums[mid]) {
                    if (target > nums[r]) {
                        r = mid - 1;
                    } else {
                        l = mid + 1;   
                    }
                } else {
                    
                    l = mid + 1;
                }
            } else {
                if (target < nums[mid]) {
                    r = mid - 1 ;
                } else {
                    // target > nums[mid]
                    if (target > nums[r]) {
                        r = mid - 1;
                    } else {
                        l = mid + 1;
                    }
                }
            }            
        }
        return false;
    }
    
    
};

看到的最好的解法:

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int left = 0, right =  nums.size()-1, mid;
        
        while(left<=right)
        {
            mid = (left + right) >> 1;
            if(nums[mid] == target) return true;

            // the only difference from the first one, trickly case, just updat left and right
            if( (nums[left] == nums[mid]) && (nums[right] == nums[mid]) ) {++left; --right;}

            else if(nums[left] <= nums[mid])
            {
                if( (nums[left]<=target) && (nums[mid] > target) ) right = mid-1;
                else left = mid + 1; 
            }
            else
            {
                if((nums[mid] < target) &&  (nums[right] >= target) ) left = mid+1;
                else right = mid-1;
            }
        }
        return false;
    }
};