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  • 137. Single Number II

    Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

    Note:

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

    Example 1:

    Input: [2,2,3,2]
    Output: 3
    

    Example 2:

    Input: [0,1,0,1,0,1,99]
    Output: 99

    reference: https://leetcode.com/problems/single-number-ii/discuss/43302/Accepted-code-with-proper-Explaination.-Does-anyone-have-a-better-idea

    对于每一个元素,统计其出现次数,若出现次数达到3,则置为0,最后所有元素的次数都为0,除了只出现一次的元素是1

    即0->1->2->0,如果用二进制表示:00->01->10->00。换一种方便的表示方法:00->10->01->00,这样可以用两个参数ones twos来表示,最后返回ones就是只出现一次的数字

    ones和twos的变化过程:

      0     0   (初始)

    0->1  0->0

    1->0  0->1

    0->0  1->0

    遍历整个数组,对每一个元素,先存入ones,再清空ones存入twos,再清空twos

    time: O(n), space: O(1)

    class Solution {
        public int singleNumber(int[] nums) {
            int ones = 0, twos = 0;
            for(int k : nums) {
                ones = (ones ^ k) & ~twos;
                twos = (twos ^ k) & ~ones;
            }
            return ones; 
        }
    }

    另一种理解方法:模拟三进制

    class Solution {
        public int singleNumber(int[] nums) {
            int one = 0, two = 0, three = 0;
            for(int i = 0; i < nums.length; i++) {
                two |= one & nums[i];
                one ^= nums[i];
                three = one & two;
                one &= ~three;
                two &= ~three;
            }
            return one;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10249861.html
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