leetcode(33) - Search in Rotated Sorted Array

int find(int* nums, int start, int end, int target) {
    if (start>end) return -1;
    int mid=0;
    
    while(start <= end){
        mid=(start+end)/2;

        if (nums[mid] < target){
            start=mid+1;
        } else {
            if (nums[mid] > target){
                end=mid-1;
            } else {
                return mid;
            }
        }
    }
    
    return -1;
    
    
    
}

int search(int* nums, int numsSize, int target) {
        int start=0, end=numsSize-1;
        int index=0; //翻转的临界点
        int num1=-1, num2=-1;
        int flag=0;
        while(index+1 < numsSize){
            
            if(nums[index] >= nums[index+1]){
                flag=1;
                break;
            }
            index++;
        }
        if(flag==0) index=0;
        printf("%d", index);
        
            if(target <= nums[index] && target >= nums[0]){
                    return find(nums, 0, index, target);
            } else {
                
                
                    return find(nums, index+1, numsSize-1, target);

            }
}

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.

本题是二分查找的变形题，主要就是找到翻转的那个临界点（下一个数值比它小）就可以了，然后看target的大小判断应该在哪个范围内进行二分查找。