剑指offer面试题11

学到了

//比照书上敲的代码，数据比较难看
int minArray(int* numbers, int numbersSize){
    if(numbersSize <= 0 || numbers == NULL)//invalid input
        exit(-1);
    int left = 0;
    int right = numbersSize-1;
    int mid = (left+right)/2;
    /*考虑单调数组*/
    if(numbers[right] > numbers[left])
        return numbers[left];
    /*非纯单调数组*/
    while(left+1 < right)
    {
        if(numbers[left] == numbers[right] && numbers[left] == numbers[mid])
            return findMin(numbers,left,right);//无法判定前后，遍历查找
        if(numbers[mid] >= numbers[left])//排除遍历的那种情况，必在左增
        {
            left = mid;//在左半增区间
        }
        else
        {
            right = mid;//在右半增区间
        }
        mid = (right+left)/2;
    }
    return numbers[right];
}
int findMin(int* numbers,int left,int right)
{
    //哈哈，正道的光，图书馆旁的广告音响终于停了哈哈哈哈哈
    int min = numbers[left];
    for(int i = left + 1;i <= right;i++)
    {
        if(min > numbers[i])
            min = numbers[i];
    }
    return min;
}

接下来是研究题解的思路
思路还是二分，不过有一说一，代码短了不少

int minArray(int* numbers, int numbersSize){
    if(!numbers || numbersSize <= 0)//invalid input
        exit(-1);
    int low = 0;
    int high = numbersSize-1;
    int mid;
    while(low < high)
    {
        // mid = (low+high)/2;
        mid = low + (high-low)/2;//放在low和high过大时发生溢出
        if(numbers[mid] > numbers[high])
        {
            low = mid+1;//mid一定不是最小的啦所以加一
        }
        else if(numbers[mid] < numbers[high])
        {
            high = mid;
        }
        else
        {
            --high;
        }
    }
    return numbers[low];
}