Interview Classic 150 Questions P33 Search in Rotated Sorted Array

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Timeline

Timeline

2025-12-04

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Divide and Conquer

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Problem:

P33 Search in Rotated Sorted Array

https://leetcode.cn/problems/search-in-rotated-sorted-array/description/?envType=study-plan-v2&envId=top-interview-150

The essence of binary search is to rule out the other half:

• ifleft halfis increasing and target is not in this interval, then target must be in theright half
• ifright halfis increasing and target is not in this interval, then target must be in theleft half

#include <vector>
using std::vector;

class Solution {
    public:
        int search(vector<int> &nums, int target)
        {
                int n = nums.size();
                int left = 0, right = n - 1;
                int mid;
                while (left <= right) {
                        mid = left + (right - left) / 2;

                        if (nums[mid] == target) {
                                return mid;
                        }

                        if (nums[left] <= nums[mid]) {
                                if (nums[left] <= target &&
                                    target < nums[mid]) {
                                        right = mid - 1;
                                } else {
                                        left = mid + 1;
                                }
                        }else if (nums[right] >= nums[mid]) {
                                if (nums[mid] < target &&
                                    target <= nums[right]) {
                                        left = mid + 1;
                                } else {
                                        right = mid - 1;
                                }
                        }
                }
                return -1;
        }
};

leetcode hot 100 rewrite

#include <vector>
using std::vector;

class Solution {
    public:
        int search(vector<int> &nums, int target)
        {
                // 1 <= nums.length <= 5000
                // -104 <= nums[i] <= 104
                // Every value in nums is unique.
                // The problem data guarantees that nums was rotated at some previously unknown index.
                // -104 <= target <= 104

                int n = nums.size();

                int left = 0, right = n - 1, mid;

                while (left <= right) {
                        mid = left + (right - left) / 2;

                        if (target == nums[mid])
                                return mid;

                        if (nums[mid] < nums[right]) {

                                if (target > nums[mid] && target <= nums[right]) // on the left side of the ascending segment
                                        left = mid + 1;
                                else
                                        right = mid - 1;

                        } else { // in the reversed segment
                                if (target < nums[mid] && target >= nums[left]) // on the left side of the reversed segment
                                        right = mid - 1;
                                else
                                        left = mid + 1;
                        }
                }

                return -1;
        }
};