[leetcode-81]Search in Rotated Sorted Array II（C)

问题描述：
Follow up for “Search in Rotated Sorted Array”:
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

分析：
根据这道题的特点，需要找到转折点，其条件是nums[i+1]

bool search(int* nums, int numsSize, int target) {
    int i;
    for(i = 0;i<numsSize;i++){
        if(nums[i]==target)
            return true;
        if(i+1<numsSize && nums[i+1]<nums[i])
            break;
    }
    int left = i+1;
    int right = numsSize-1;

    while(left<=right){
        int mid = (left+right)/2;
        if(nums[mid]==target)
            return true;
        else if(nums[mid]<target)
            left = mid+1;
        else
            right = mid-1;
    }

    return false;
}

代码2：暴力搜索（8ms)

int search(int* nums, int numsSize, int target) {
    int i;
    for(i = 0;i<numsSize;i++){
        if(nums[i]==target)
            break;
    }
    if(i>=numsSize)
        return -1;

    return i;
}