【LeetCode】#33搜索旋转排序数组(Search in Rotated Sorted Array)
题目描述
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
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
Description
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
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
解法
class Solution {
public int search(int[] nums, int target) {
if(nums.length<=0){
return -1;
}
int len = nums.length, l=0, m=len/2, r=len-1;
while(true){
int left=nums[l], mid=nums[m], right=nums[r];
if(target==left){
return l;
}
if(target==mid){
return m;
}
if(target==right){
return r;
}
if(Math.abs(m-l)<=1 && Math.abs(m-r)<=1){
return -1;
}
if(left<mid){
if(left<target && target<mid){
r = m;
m = (l+r)/2;
continue;
}else{
l = m;
m = (l+r)/2;
continue;
}
}
if(mid<right){
if(mid<target && target<right){
l = m;
m = (l+r)/2;
continue;
}else{
r = m;
m = (l+r)/2;
continue;
}
}
}
}
}