题目

二分查找

使用两次二分查找，一次查找 target 的开始位置，一次查找结束位置，时间复杂度 $O(\log n)$ ，空间复杂度 $O(1)$ ：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int n = nums.size();
        if(n == 0)
            return vector<int>{-1, -1};
        if(n == 1)
            return nums[0] == target ? vector<int>{0, 0} : vector<int>{-1, -1};
        int l = 0, r = n - 1, midl = -1, midr = -1, tmp;
        while(l <= r) {
            tmp = (l + r) / 2;
            if(nums[tmp] == target) {
                midl = tmp;
                r = tmp - 1;
            }
            else if(target < nums[tmp])
                r = tmp - 1;
            else l = tmp + 1;
        }
        l = 0;
        r = n - 1;
        while(l <= r) {
            tmp = (l + r) / 2;
            if(nums[tmp] == target) {
                midr = tmp;
                l = tmp + 1;
            }
            else if(target < nums[tmp])
                r = tmp - 1;
            else l = tmp + 1;
        }
        
        return vector<int>{midl, midr};
    }
};
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