题目
给定一个按照升序排列的整数数组 nums, 和一个目标值 target. 找出给定目标值在数组中的开始位置和结束位置.
你的算法时间复杂度必须是? O(log n) 级别.
如果数组中不存在目标值, 返回?[-1, -1].
示例 1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]
示例? 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]
本题同 [剑指 Offer] 面试题 53 - I. 在排序数组中查找数字 I https://www.cnblogs.com/galaxy-hao/p/12669615.html
思路一: 二分查找
代码
时间复杂度: O(logn)
空间复杂度: O(1)
- class Solution {
- public:
- vector<int> searchRange(vector<int>& nums, int target) {
- if (nums.empty()) return {-1, -1};
- int left = searchLeft(nums, target);
- int right = searchRight(nums, target);
- return {left, right};
- }
- int searchLeft(vector<int> &nums, int target) {
- int left = 0, right = nums.size() - 1;
- while (left <= right) {
- int mid = left + (right - left) / 2;
- if (nums[mid] == target) {
- if (mid == 0 || (mid - 1>= 0 && nums[mid - 1] != target)) {
- return mid;
- }
- right = mid - 1;
- } else if (nums[mid] <target) {
- left = mid + 1;
- } else {
- right = mid - 1;
- }
- }
- return -1;
- }
- int searchRight(vector<int> &nums, int target) {
- int left = 0, right = nums.size() - 1;
- while (left <= right) {
- int mid = left + (right - left) / 2;
- if (nums[mid] == target) {
- if (mid == nums.size() - 1 || (mid + 1 <= nums.size() - 1 && nums[mid + 1] != target)) {
- return mid;
- }
- left = mid + 1;
- } else if (nums[mid] <target) {
- left = mid + 1;
- } else {
- right = mid - 1;
- }
- }
- return -1;
- }
- };
另一种写法
- class Solution {
- public:
- vector<int> searchRange(vector<int>& nums, int target) {
- if (nums.empty()) return {-1, -1};
- int left = searchLeft(nums, target);
- int right = searchRight(nums, target);
- return {left, right};
- }
- int searchLeft(vector<int> &nums, int target) {
- int left = 0, right = nums.size() - 1;
- while (left <= right) {
- int mid = left + (right - left) / 2;
- if (nums[mid] == target) {
- // 一直向左找
- while (mid - 1>= 0 && nums[mid - 1] == target) {
- --mid;
- }
- return mid;
- } else if (nums[mid] <target) {
- left = mid + 1;
- } else {
- right = mid - 1;
- }
- }
- return -1;
- }
- int searchRight(vector<int> &nums, int target) {
- int left = 0, right = nums.size() - 1;
- while (left <= right) {
- int mid = left + (right - left) / 2;
- if (nums[mid] == target) {
- // 一直向右找
- while (mid + 1 <= nums.size() - 1 && nums[mid + 1] == target) {
- ++mid;
- }
- return mid;
- } else if (nums[mid] <target) {
- left = mid + 1;
- } else {
- right = mid - 1;
- }
- }
- return -1;
- }
- };
思路二: STL
lower_bound: 返回一个迭代器, 指向键值 >= key 的第一个元素
upper_bound: 返回一个迭代器, 指向键值 > key 的第一个元素
代码
时间复杂度: O(logn)
空间复杂度: O(1)
- class Solution {
- public:
- vector<int> searchRange(vector<int>& nums, int target) {
- if (nums.empty()) return {-1, -1};
- auto left = lower_bound(nums.begin(), nums.end(), target);
- auto right = upper_bound(nums.begin(), nums.end(), target);
- if (left == right) return {-1, -1};
- return {left - nums.begin(), right - nums.begin() - 1};
- }
- };
来源: http://www.bubuko.com/infodetail-3499948.html