分析

• 本题的考点：二分。

• 二次二分即可，第一次二分出左端点，第二次二分出右端点。

代码

• C++

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {

        if (nums.empty()) return {-1, -1};

        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = l + r >> 1;
            if (nums[mid] >= target) r = mid;
            else l = mid + 1;
        }

        if (nums[r] != target) return {-1, -1};

        int L = l;
        l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (nums[mid] <= target) l = mid;
            else r = mid - 1;
        }

        return {L, r};
    }
};

• Java

class Solution {
    public int[] searchRange(int[] nums, int target) {

        if (nums.length == 0) return new int[]{-1, -1};

        int l = 0, r = nums.length - 1;
        while (l < r) {
            int mid = l + r >> 1;
            if (nums[mid] >= target) r = mid;
            else l = mid + 1;
        }

        if (nums[r] != target) return new int[]{-1, -1};

        int L = l;
        l = 0; r = nums.length - 1;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (nums[mid] <= target) l = mid;
            else r = mid - 1;
        }

        return new int[]{L, r};
    }
}

时空复杂度分析

• 时间复杂度：$O(log(n)$，n为数组长度。

• 空间复杂度：$O(1)$。