小旭讲解 LeetCode 33. Search in Rotated Sorted Array

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原题

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:

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

Example 2:

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

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代码

class Solution {
    public:
    int findP(vector<int>& nums) {
        if (nums[0] < nums[nums.size() - 1]) return -1;
        int l = 0, h = nums.size() - 1, m = 0;
        while (l < h) {
            m = l + (h - l + 1) / 2;
            if (nums[m] < nums[0]) {
                h = m - 1;
                // h = m;
            } else {
                l = m;
                // l = m + 1;
            }
        }
        return l;
    }
    int findT(vector<int>& nums, int l, int h, int t) {
        int m = 0;
        while (l < h) {
            m = l + (h - l) / 2;
            if (nums[m] < t) {
                l = m + 1;
            } else {
                h = m;
            }
        }
        return l;
    }
    int search(vector<int>& nums, int target) {
        if (nums.size() == 0) return -1;
        if (nums.size() == 1) return nums[0] == target ? 0 : -1;
        int p = findP(nums);
        if (p == -1) {
            p = findT(nums, 0 , nums.size() - 1, target);
        } else if (nums[0] <= target) {
            p = findT(nums, 0, p, target);
        } else {
            p = findT(nums, p + 1, nums.size() - 1, target);
        }
        return nums[p] == target ? p : -1;
    }
};

复杂度分析

时间复杂度：O(\log n)
空间复杂度：O(1)

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