540. Single element in ordered array

>540. A single element in an ordered array <

>singleNonDuplicate<

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One 、 Their thinking

1、 Solution 1 ( Java )

Solution ideas ： XOR pairing + Two points search
adopt An operation Medium mid ^ 1 Exclusive or Operator to unify parity

1. i For even when ：i ^ 1 == i + 1, The result is the odd number after the original even number plus one
2. i In an odd number of ：i ^ 1 == i - 1, The result is the even number after subtracting one from the original odd number

/** * @author listen1024 * @date 2022-2-14 * 540.  A single element in an ordered array  */
class Solution {
    
    public int singleNonDuplicate(int[] nums) {
    
        int low = 0;
        int high = nums.length - 1;
        while (low < high) {
    
            int mid = (low + high) /2;
            if (nums[mid] == nums[mid ^ 1]){
    
                low = mid + 1;
            } else {
    
                high = mid;
            }
        }
        return nums[low];
    }
}

The screenshot of the running result is as follows ：