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  • 13 Search in Rotated Sorted Array II

    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

    这道题的核心是用二分法,间或应用一些其他的技巧,可以说是二分法的进阶版。。。

    class Solution {
    public:
        int search(vector<int>& nums, int target) {
            
            
            int left = 0;
            int right = nums.size() - 1;
            
            while(left <= right)
            {
                int mid = left  +  (right - left)/2;
                
                if(nums[mid] == target) return mid;
                
                if(nums[mid] < nums[right])
                {
                    if(target > nums[mid] && target <= nums[right]) left = mid + 1;
                    else right = mid - 1;
                    
                }
                else
                {
                   if(target > nums[mid] || target <= nums[right])  left = mid + 1;
                   else right = mid - 1;
                }
            }
            
            return -1;
            
        }
        
        
    };
    
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  • 原文地址:https://www.cnblogs.com/xiaoyisun06/p/11380290.html
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