Karl's Nest
0%

leetcode-day-19

Posted on In
Day-19 Search in Rotated Sorted Array

LeetCode 30 days Challenge - Day 19

本系列将对LeetCode新推出的30天算法挑战进行总结记录，旨在记录学习成果、方便未来查阅，同时望为广大网友提供帮助。

Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

1
2
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

1
2
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Solution

题目要求分析：给定一个经过“旋转”的升序排列数组，在其中查找target，若不存在返回-1。

解法：

本题有多种解法，作者选用的是一种比较直观的方法，其他更为“tricky”的题解可参考原题讨论区。

思路大体上分为两步：

  1. 二分搜索，找到原升序排列数组第一个元素在给定数组中的位置sep。
  2. 通过映射，将给定数组的元素位置“映射”到实际位置：true_pos = (pos + sep) % nums.size()
  3. 在此基础上，进行二分搜索。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
int search(vector<int>& nums, int target) {
    int l = 0, r = nums.size()-1, sep = 0;
    while (l < r) {
        int mid = (l + r) / 2;
        if (nums[mid] > nums[r]) l = mid + 1;
        else r = mid;
    }
    sep = l;
    l = 0;
    r = nums.size()-1;
    while (l <= r) {
        int mid = (l + r) / 2;
        int mapped_mid = (mid + sep) % nums.size();
        if (nums[mapped_mid] == target) return mapped_mid;
        else if (nums[mapped_mid] < target) l = mid + 1;
        else r = mid - 1;
    }
    return -1;
}

传送门：Search in Rotated Sorted Array

Karl