成都网站开发费用,做行程的网站,网站建设的请示,从58做网站怎么做Problem: 34. 在排序数组中查找元素的第一个和最后一个位置 文章目录 题目描述思路复杂度Code 题目描述 思路 Problem: 二分查找常用解题模板#xff08;带一道leetcode题目#xff09; 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 ( l o g n )… Problem: 34. 在排序数组中查找元素的第一个和最后一个位置 文章目录 题目描述思路复杂度Code 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 ( l o g n ) O(logn) O(logn)其中 n n n为数组nums的大小 空间复杂度: O ( n ) O(n) 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 vectorint*/vectorint searchRange(vectorint nums, int target) {if (nums.size() 0) {return {-1, -1};}vectorint 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(vectorint 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 boundsif (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(vectorint 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 boundsif (right 0 || nums[right] ! target) {return -1;}return right;}
};
