Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,

Given [5, 7, 7, 8, 8, 10] andtarget value 8,

return [3, 4].

 

思路：本题还是在考察二分查找法的使用。首先使用二分查找找到一个target的位置，该位置将数组分为两个子数组，左边的子数组都是小于等于target，右边的子数组都是大于等于target。可以再利用二分查找，分别找到target的边界。代码如下：

/**
 * Return an arrayof size *returnSize.
 * Note: Thereturned array must be malloced, assume caller calls free().
 */
int* searchRange(int* nums, int numsSize, int target, int* returnSize)
{
    int resleft, resright;
    int left, right;
    int mid = -1;
    int resmid;
 
    left = 0; right = numsSize - 1;
    while(left <= right)
    {
        mid = left + (right-left)/2;
        if(nums[mid] == target)
        {
            break;
        }
        if(nums[mid] > target)
        {
            right -= 1;
        }
        else
        {
            left += 1;
        }
    }
 
    if(left > right)
    {
        int *res = calloc(2, sizeof(int));
        res[0] = -1;
        res[1] = -1;
        *returnSize = 2;
        return res;
    }
 
    left = 0;
    right = mid;
    while(left < right)
    {
        resmid = left + (right-left)/2;
        if(nums[resmid] < target)
        {
            left = resmid+1;
        }
        else
        {
            right = resmid;
        }
    }
    resleft = left;
 
    left = mid;
    right = numsSize - 1;
    while(left <= right)
    {
        resmid = left + (right-left)/2;
        if(nums[resmid] == target)
        {
            left = resmid+1;
        }
        else
        {
            right = resmid-1;
        }
    }
    resright = right;
 
    int *res = calloc(2, sizeof(int));
    res[0] = resleft;
    res[1] = resright;
    *returnSize = 2;
    return res;
}