给定按升序排序的整数数组，找到给定目标值的起始和终止位置。 您的算法的运行时复杂度必须是O（log n）的顺序。

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */int* searchRange(int* nums, int numsSize, int target, int* returnSize) {    int low_i,low_j,high_i,high_j;    int low_middle,high_middle;    low_i = 0;    low_j = numsSize-1;    *returnSize=2;    int *res = (int*)calloc(2, sizeof(int));    res[0] = -1;    res[1] = -1;    while(low_i<=low_j)    {        low_middle = (low_i+low_j)/2;        if(nums[low_middle]==target)        {            if(low_middle==0) break;            else            {                if(nums[low_middle-1]target) break;                else high_i = high_middle+1;            }        }        else            high_j = high_middle-1;    }    res[0] = low_middle;    res[1] = high_middle;    return res;    }

思路：使用二分查找，查找目标值的低位下标和高位下标。设置4个变量，代表查找低位下标和高位下标的起始位置。不断二分查找，最终找到目标！

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */int* searchRange(int* nums, int numsSize, int target, int* returnSize) {    int low_i,low_j,high_i,high_j;    int low_middle,high_middle;    low_i = 0;    low_j = numsSize-1;    *returnSize=2;    int *res = (int*)calloc(2, sizeof(int));    res[0] = -1;    res[1] = -1;    while(low_i<=low_j)    {        low_middle = (low_i+low_j)/2;        if(nums[low_middle]==target)        {            if(low_middle==0) break;            else            {                if(nums[low_middle-1]<target) break;                else low_j = low_middle;            }        }        else if(nums[low_middle]<target)            low_i = low_middle+1;        else            low_j = low_middle-1;    }    if(nums[low_middle]!=target)         return res;    high_i = low_middle;    high_j = numsSize-1;    while(high_i<=high_j)    {        high_middle = (high_i+high_j)/2;        if(nums[high_middle]==target)        {            if(high_middle==numsSize-1) break;            else            {                if(nums[high_middle+1]>target) break;                else high_i = high_middle+1;            }        }        else            high_j = high_middle-1;    }    res[0] = low_middle;    res[1] = high_middle;    return res;    }