540. Single Element in a Sorted Array

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问题描述:

Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.

Example 1:

Input: [1,1,2,3,3,4,4,8,8]
Output: 2

 

Example 2:

Input: [3,3,7,7,10,11,11]
Output: 10

 

Note: Your solution should run in O(log n) time and O(1) space.

 

解题思路:

要求我们要在O(logn)时间和O(1)空间内解决,首先想到二分搜索。

在二分搜索的应用中,如何找到移动的条件是十分关键的。

 

我们先观察一下:

值:  1 1 2 2 4 4 5 5 9

下标: 0 1 2 3 4 5 6 7 8

此时中值为4,下标也为4。

若从头到尾都是成双成对,那偶数位的应该等于它后面的奇数位。

如果不等于,那么说明前半段出现了一个奇数。

若中值位一个奇数,若它与前面的数字相同,则说明前面都是成双成对,只有一个的数字是出现在后面。

同时我们也需要判断当前值是否是我们要找的值

即不等于左边的值,又不等于右边的值

if((nums[mid] != l_Num || mid == 0) && (nums[mid] != r_Num || mid == n-1)){
                return nums[mid];
            }

在这里需要注意的是数组头部和尾部的判断,否则会出现死循环

 

代码:

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int n = nums.size();
        if(n == 1)
            return nums[0];
        int left = 0;
        int right = n-1;
        int mid;
        while(left <= right){
            mid = left + (right - left)/2;
            int l_Num = mid-1 > -1 ? nums[mid-1] : nums[mid];
            int r_Num = mid + 1 < n ? nums[mid+1] : nums[mid];
            if((nums[mid] != l_Num || mid == 0) && (nums[mid] != r_Num || mid == n-1)){
                return nums[mid];
            }
            if(mid % 2 == 1){
                if(l_Num == nums[mid]){
                    left = mid+1;
                }else{
                    right = mid;
                }
            }else{
                if(l_Num == nums[mid]){
                    right = mid;
                }else{
                    left = mid+1;
                }
            }
        }
        return nums[mid];
    }
};

 

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