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leetcode第33题--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] and target value 8,
return [3, 4].

给的那个一个有序数组和一个目标target,找到这个目标在有序数组里的开始和结束的下标。如果不存在就返回 -1 -1。注,如果数组中目标值只有一个,那就返回两个相同的下标。

用binary search找到target   然后再在左边binary search找到左边(包括mid)第一个目标,记录下标。再在右边binary 找到最后一个目标。记录下标。返回。

代码如下:

class Solution {public:vector<int> searchRange(int A[], int n, int target){    vector<int> wr, ans;    wr.push_back(-1); // 要学会可以用 vector<int> wr(2,-1);直接将其初始化为想要的    wr.push_back(-1);    if(n == 0)    return wr;    int l = 0, r = n -1, mid = 0;    while(l <= r)    {        mid = (l+r)/2;        if (A[mid] < target) // 如果mid小于target那么可以把左边一块去掉        {            l = mid + 1;            continue;        }        if (A[mid] > target) // 如果mid大于target那么可以把右边一块去掉        {            r = mid -1;            continue;        }        if (A[mid] == target)// mid 找到了刚好,那就肯定在这里面要有return,如果一直没有找到,跳出while后就返回wr        {            // find start from l to mid,binary search            int start = l, l_r = mid; // l_r指mid左边的最右            while(l <= l_r)            {                int tmp_mid = (l + l_r)/2;                if(A[tmp_mid] != target)                    {l = tmp_mid + 1;}                if(A[tmp_mid] == target)                    {l_r = tmp_mid - 1; start = tmp_mid;}            }            ans.push_back(start);            // find end from mid to r,binary search            int r_l = mid, End = mid; // r_l指mid右边的最左            while(r_l <= r)            {                int tmp_mid = (r_l + r)/2;                if(A[tmp_mid] != target)                    {r = tmp_mid - 1;}                if(A[tmp_mid] == target)                    {r_l = tmp_mid + 1; End = tmp_mid;}            }            ans.push_back(End);            return ans;        }    }    return wr;   }};

 

leetcode第33题--Search for a Range