面试经典150题 P33 搜索旋转排序数组

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时间轴

时间轴

2025-12-04

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分治

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题目：

P33 搜索旋转排序数组

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

二分本质是排除掉另一半：

• 如果左半边是递增的且 target 不处于这个区间，那么 target 必定在右半边
• 如果右半边是递增的且 target 不处于这个区间，那么 target 必定在左半边

#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
                // nums 中的每个值都 独一无二
                // 题目数据保证 nums 在预先未知的某个下标上进行了旋转
                // -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]) // 在正序段左侧
                                        left = mid + 1;
                                else
                                        right = mid - 1;

                        } else { // 在逆序段
                                if (target < nums[mid] && target >= nums[left]) // 在逆序段左侧
                                        right = mid - 1;
                                else
                                        left = mid + 1;
                        }
                }

                return -1;
        }
};