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  • Leetcode: 33. Search in Rotated Sorted Array

    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.

    思路

    • 二分查找
    • 有两个关键,由于数组经过旋转过,所以应该先判断mid落在前半段还是后半段
    • 然后再根据target的大小判断怎么缩小范围

    代码

    class Solution {
    public:
       int search(vector<int>& nums, int target) {
    		int len = nums.size();
    		if (len == 0) return -1;
    		
    		int low = 0, high = len - 1, mid = 0;
    		while (low <= high){
    			mid = low + (high - low) / 2;
    
    			if (nums[mid] == target) return mid;
    
                //落在前半段
    			if (nums[mid] >= nums[low]){
    			    //落在前半段的前半段
    				if (nums[low] <= target && target < nums[mid])
    					high = mid - 1;
    				else low = mid + 1;
    				
    			}
    			else{
    			    //落在后半段的后半段
    				if (nums[mid] < target && target <= nums[high])
    					low = mid + 1;
    				else high = mid - 1;
    			}
    		}
    
    		return -1;
    	}
    };
    
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  • 原文地址:https://www.cnblogs.com/lengender-12/p/6835217.html
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