【leetcode】#数组【Python】33. Search in Rotated Sorted Array 搜索旋转排序数组

链接：

https://leetcode-cn.com/problems/search-in-rotated-sorted-array/

题目：

假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。
你可以假设数组中不存在重复的元素。你的算法时间复杂度必须是 O(log n) 级别。
示例 1:
输入: nums = [4,5,6,7,0,1,2], target = 0
输出: 4
示例 2:
输入: nums = [4,5,6,7,0,1,2], target = 3
输出: -1

我的解法：（28ms，击败92%）先判断当前mid的位置，分别做判断。
比方[5671234]，tar=2。 mid = 1，此时，如果target在1-4之间，取右半边；target小于1或者大于4在左边。写得挺繁琐的。。

class Solution:
    def search(self, nums, target):
        if nums == []:
            return -1
        left = 0
        right = len(nums) - 1
        while left <= right:           
            mid = int((left + right)/2)
            if target == nums[mid]:
                return mid
            if nums[left] > nums[mid]:
                if target > nums[mid] and target <= nums[right]:
                    left = mid + 1
                elif target < nums[mid] or target > nums[right]:
                    right = mid - 1
            elif nums[left] < nums[mid]:
                if target >= nums[left] and target < nums[mid]:
                    right = mid - 1
                elif target > nums[mid] or target < nums[left]:
                    left = mid + 1
            else:
                if nums[right] == target:
                    return right
                else:
                    return -1

别人的解法：思路类似，精简版，优秀

class Solution:
    def search(self, nums, target):
        if not nums:
            return -1
        low, high = 0, len(nums) - 1
        while low <= high:
            mid = (low + high) / 2
            if target == nums[mid]:
                return mid
            # 左边的序列是正序排列的
            if nums[low] <= nums[mid]:
            	# target在前半部分正排序内
                if nums[low] <= target <= nums[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            # 左边是先增后小的，说明后半部分是正序排列
            else:
            	# 如果target在后半部分排列内
                if nums[mid] <= target <= nums[high]:
                    low = mid + 1
                else:
                    high = mid - 1
        return -1

别人的解法2: 计算mid的时候，不同情况下赋值成inf，或者-inf；就可以直接做二分查找了。

Let’s say nums looks like this: [12, 13, 14, 15, 16, 17, 18, 19, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], Because it’s not fully sorted, we can’t do normal binary search. But here comes the trick:

If target is let’s say 14, then we adjust nums to this, where “inf” means infinity:
[12, 13, 14, 15, 16, 17, 18, 19, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf]
If target is let’s say 7, then we adjust nums to this:
[-inf, -inf, -inf, -inf, -inf, -inf, -inf, -inf, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
And then we can simply do ordinary binary search.

int search(vector<int>& nums, int target) {
     
    int lo = 0, hi = nums.size();
    while (lo < hi) {
     
        int mid = (lo + hi) / 2;
        
        double num = (nums[mid] < nums[0]) == (target < nums[0])
                   ? nums[mid]
                   : target < nums[0] ? -INFINITY : INFINITY;
                   
        if (num < target)
            lo = mid + 1;
        else if (num > target)
            hi = mid;
        else
            return mid;
    }
    return -1;
}

python版本：新思路

class Solution:
    def search(self, nums, target):
        if not nums:
            return -1
        low, high = 0, len(nums) - 1
        while low <= high:
            mid = int((low + high) / 2)
            #num_mid = nums[mid]
            if target < nums[low] and nums[mid] >= nums[low]:
                nums[mid] = - float("inf")
            elif target >= nums[low] and nums[mid] < nums[low]:
                nums[mid] = float("inf")
            
            if target < nums[mid]:
                high = mid - 1
            elif target > nums[mid]:
                low = mid + 1
            else:
                return mid
        return -1