码迷,mamicode.com
首页 > 编程语言 > 详细

【LeetCode】34. 在排序数组中查找元素的第一个和最后一个位置

时间:2020-04-09 21:29:54      阅读:89      评论:0      收藏:0      [点我收藏+]

标签:off   时间   面试   查找   排序   offer   else   galaxy   ==   

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

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};
    }
};

【LeetCode】34. 在排序数组中查找元素的第一个和最后一个位置

标签:off   时间   面试   查找   排序   offer   else   galaxy   ==   

原文地址:https://www.cnblogs.com/galaxy-hao/p/12669640.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!