Java for LeetCode 137 Single Number II

標籤：

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

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

解題思路一：

一個int的長度是32，因此可以開一個長度為32的數組，表示nums中所有元素各位1的個數和，然後%3即可得到結果。

JAVA實現如下：

public int singleNumber(int[] nums) {int[] bitnum = new int[32];int res = 0;for (int i = 0; i < 32; i++) {for (int j = 0; j < nums.length; j++)bitnum[i] += (nums[j] >> i) & 1;res += (bitnum[i] % 3) << i;}return res;}

 解題思路二：

分別用三個變數bit0、bit1、bit2表示nums元素中1個數為0、1、2的分布，最後返回bit1即可，JAVA實現如下：

public int singleNumber(int[] nums) {int bit0 = ~0, bit1 = 0, bit2 = 0, oldTwo;for (int i = 0; i < nums.length; i++) {oldTwo = bit2;bit2 = (bit1 & nums[i]) | (bit2 & ~nums[i]);bit1 = (bit0 & nums[i]) | (bit1 & ~nums[i]);bit0 = (oldTwo & nums[i]) | (bit0 & ~nums[i]);}return bit1;}

 

Java for LeetCode 137 Single Number II