leetcode算法题--有序数组中的单一元素

原题链接：https://leetcode-cn.com/problems/single-element-in-a-sorted-array/

使用二分查找法，索引是偶数则应后面一个相等，索引为奇数则应与前面一个相等，逐步缩小范围

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int l = 0, h = nums.size() - 1;
        while (l < h) {
            int m = (h - l) / 2 + l;
            if (nums[m] == nums[m ^ 1]) {
                l = m + 1;
            } else {
                h = m;
            }
        }
        return nums[l];
    }
};

单一元素的索引必为偶数，于是使用二分查找，逐步缩小范围，直到找到数

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int l = 0, h = nums.size() - 1;
        while (l < h) {
            int m = (h - l) / 2 + l;
            cout << " h is " << h << " l is " << l << " m is " << m << endl;
            m -= m & 1;
            if (nums[m] == nums[m + 1]) {
                l = m + 2;
            } else {
                h = m;                
            }
        }
        return nums[l];
    }
};