Problem: 34. 在排序数组中查找元素的第一个和最后一个位置

题目描述

思路

Problem: 二分查找常用解题模板（带一道leetcode题目）

直接套用上述中的寻找左、右边界的二分查找模板即可

复杂度

时间复杂度:

$O(logn)$；其中$n$为数组nums的大小

空间复杂度:

$O(n)$

Code

class Solution {
public:
    /**
     * Finds the first and last position of an element in a sorted array
     *
     * @param nums Given array
     * @param target Given target number
     * @return vector<int>
     */
    vector<int> searchRange(vector<int>& nums, int target) {
        if (nums.size() == 0) {
            return {-1, -1};
        }
        vector<int> res(2);
        res[0] = left_bound(nums, target);
        res[1] = right_bound(nums, target);
        return res;
    }

    /**
     * Queries the left boundary for a number less than the specified number
     *
     * @param nums Given array
     * @param target Given target number
     * @return int
     */
    int left_bound(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                right = mid - 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else if (nums[mid] < target) {
                left = mid + 1;
            }
        }
        // Check out of bounds
        if (left >= nums.size() || nums[left] != target) {
            return -1;
        }
        return left;
    }

    /**
     * Queries the right boundary for a number less than the specified number
     * 
     * @param nums Given array
     * @param target Given target number
     * @return int
     */
    int right_bound(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else if (nums[mid] < target) {
                left = mid + 1;
            }
        }
        // Check out of bounds
        if (right < 0 || nums[right] != target) {
            return -1;
        }
        return right;
    }
};