81. Search in Rotated Sorted Array II

谁借莪１个温暖的怀抱￠ 2022-03-07 14:45 177阅读 0赞

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.  
(i.e., \[0,0,1,2,2,5,6\] might become \[2,5,6,0,0,1,2\]).  
You are given a target value to search. If found in the array return true, otherwise return false.  
Example 1:

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

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
    Output: false

Follow up:

This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
    Would this affect the run-time complexity? How and why?

难度：medium

题目：假设一个按升序排序的数组在某个未知的轴上旋转。搜索给定目标值，如果找到返回true，否则返回false。

思路：二叉搜索。  
分4种情况。

case 1：从头到尾升序。
    case 2：从头到尾降序。
    case 3：升序的元素多，降序少。
    case 4：升序的元素少，降序多。

Runtime: 0 ms, faster than 100.00% of Java online submissions for Search in Rotated Sorted Array II.  
Memory Usage: 37.6 MB, less than 1.00% of Java online submissions for Search in Rotated Sorted Array II.

class Solution {
        public boolean search(int[] nums, int target) {
            int left = 0, right = nums.length - 1;
            while (left <= right) {
                int mid = left + (right - left) / 2;
                if (target == nums[mid]) {
                    return true;
                }
                
                // left < mid </> right
                if (nums[left] < nums[mid]) {
                    if (target > nums[mid] || target < nums[left]) {
                        left = mid + 1;
                    } else {
                        right = mid - 1;
                    }
                } else if (nums[left] > nums[mid]) { // left > mid </> right
                    if (target < nums[mid] || target > nums[right]) {
                        right = mid - 1;
                    } else {
                        left = mid + 1;
                    }
                } else { // left = mid
                    left += 1;
                }
            }
            
            return false;
        }
    }