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
思路:
要求时间复杂度必须是 O(log n),所以需要用二分查找法.
总共有以下几种情况:一.nums总长度为0,二.数组为升序排列,三.数组有旋转:1.target存在则只有可能出现在大数端;2.target存在则只有可能出现在小数端
当数组有旋转,且出现在大端时,注意将范围缩小到大端做法:(nums[hight] <= nums[nums_len-1]) && (nums[(low+hight)/2] < nums[hight])//表明中间点还在小的一端
代码:
//方法一:要求时间复杂度必须是 O(log n),所以需要用二分查找法.
//总共有以下几种情况:一.nums总长度为0,二.数组为升序排列,三.数组有旋转:1.target存在则只有可能出现在大数端;2.target存在则只有可能出现在小数端
//当数组有旋转,且出现在大端时,注意将范围缩小到大端做法:(nums[hight] <= nums[nums_len-1]) && (nums[(low+hight)/2] < nums[hight])//表明中间点还在小的一端
class Solution {
public:
int search(vector<int>& nums, int target) {
int nums_len = nums.size();
//特殊情况
if(nums_len == 0)
return -1;
int low = 0, hight = nums_len - 1;
if(nums[nums_len - 1] >= nums[0]) //升序排列的数组
{
while(low <= hight)
{
if(nums[(low+hight)/2] == target)
return (low+hight)/2;
else if(nums[(low+hight)/2] > target)
hight = (low+hight)/2 - 1; //防止偶数个
else
low = (low+hight)/2 + 1; //防止偶数个
}
}
else if(target >= nums[0]) //表明target存在则只有可能出现在大数端
{
while(low <= hight)
{
if( (nums[hight] <= nums[nums_len-1]) && (nums[(low+hight)/2] < nums[hight]) ) //表明中间点还在小的一端
{
hight = (low+hight)/2 - 1;
}
else //表明中间点落在了大的一端,二分查找
{
if(nums[(low+hight)/2] == target)
return (low+hight)/2;
else if(nums[(low+hight)/2] > target)
hight = (low+hight)/2 - 1; //防止偶数个
else
low = (low+hight)/2 + 1; //防止偶数个
}
}
}
else //表明target存在则只有可能出现在小数端
{
while(low <= hight)
{
if(nums[(low+hight)/2] > nums[hight]) //表明中间点还在大的一端
{
low = (low+hight)/2 + 1;
}
else //表明中间点落在了小的一端,二分查找
{
if(nums[(low+hight)/2] == target)
return (low+hight)/2;
else if(nums[(low+hight)/2] < target)
low = (low+hight)/2 + 1;
else
hight = (low+hight)/2 - 1;
}
}
}
return -1;
}
};