LeetCode Day28 Search for a Range

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int left=0,right=nums.size()-1,mid;
        while(left<=right){
            mid=(left+right)/2;
            
            if(target==nums[mid]){
                int i=mid,j=mid;
                while(target==nums[i-1]&&i>0) i--;
                while(target==nums[j+1]&&j<nums.size()-1) j++;
                return {i,j};
            }
            else if(target>nums[mid]) left=mid+1;
            else right=mid-1;
        }
        return {-1,-1};
    }
};

上面的算法不是严格意义上的O(logn)的算法，因为在最坏的情况下会变成O(n)，比如当数组里的数全是目标值的话，从中间向两边找边界就会一直遍历完整个数组，那么我们下面来看一种真正意义上的O(logn)的算法，使用两次二分查找法，第一次找到左边界，第二次调用找到右边界即可，具体代码如下：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if(nums.empty()) return {-1,-1};
        vector<int> res(2, -1);
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) left = mid + 1;
            else right = mid;
        }
        if (nums[right] != target) return res;
        res[0] = right;
        right = nums.size()-1;
        while (left < right) {
            int mid =(right + left) / 2+1;//整型除法只会向下取整，现改为向上取整，保证中间值一直>left，可以一直向右检查
            if (nums[mid] == target) left = mid;//只存在>或==的情况
            else right= mid-1;
        }
        res[1] = left;
        return res;
    }
};