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136. single number 解答参考答案

admin 11月 13, 2020 0

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

解答

public class Solution {
	public int singleNumber(int[] nums) {
		int result = 0;
		Arrays.sort(nums);
		int l = nums.length - 1;
		if (l == 0) {
			result = 0;
		} else if (nums[l] != nums[l - 1] && nums[l] != nums[l - 2]) {
			result = l;
		} else if (nums[0] != nums[1]) {
			result = 0;
		} else {
			for (int i = 2; i < l; i++) {
				result = i;
				if (nums[i] != nums[i + 1] && nums[i] != nums[i - 1]) {
					break;
				}
			}
		}
		return nums[result];
	}
}

参考答案

public int singleNumber(int[] nums) {
    int ans =0;
    int len = nums.length;
    for(int i=0;i!=len;i++)
        ans ^= nums[i];
    return ans;
}