Leetccode 136 137 260 SingleNumber I II III
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Leetccode 136 SingleNumber I
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?
1.自己想到了先排序,再遍历数组找,复杂度较高
2.参考其他想法,最好的是利用异或,详见代码
import java.util.Arrays; public class S136 { public int singleNumber(int[] nums) { //AC but not good /* Arrays.sort(nums); int i = 0; for(;i<nums.length-1;i+=2){ if(nums[i]!=nums[i+1]){ return nums[i]; } } return nums[i];*/ //best one 异或运算的神奇之处 1.a^b == b^a 2.0^a == a if(nums.length<1) return 0; int ret = nums[0]; for(int i = 0;i<nums.length;i++){ ret = ret^nums[i]; } return ret; } }
Leetccode 137 SingleNumber II
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?
1.利用排序和遍历同样可以AC
2.利用异或,详见代码
public class S137 { public int singleNumber(int[] nums) { //AC but not good /* Arrays.sort(nums); int i = 0; for(;i<nums.length-1;i+=3){ if(nums[i]!=nums[i+1]){ return nums[i]; } } return nums[i];*/ //a general algorithm int a[] = new int[32]; int ret = 0; for(int i = 0;i<32;i++){ for(int j = 0;j<nums.length;j++){ a[i] += (nums[j]>>i)&1; } ret |= (a[i]%3)<<i; } return ret; } }
Leetccode 260 SingleNumber III
Given an array of numbers nums
, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
For example:
Given nums = [1, 2, 1, 3, 2, 5]
, return [3, 5]
.
Note:
- The order of the result is not important. So in the above example,
[5, 3]
is also correct. - Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
思路:先将所有元素异或得到的结果ret肯定不为零,再移位寻找第一个不为零的二进制位,记录位置pos。再遍历数组,将所有pos位置为零的数异或得到num1,所有pos位置为一的数异或得到num2,num1和num2即answer。因为ret中不为零的二进制位所对应位肯定是num1和num2对应位异或,必定是num1和num2此位不同,将数组分为两组分别异或其实就是第一种情况的解法了。详见代码
public class S260 { public int[] singleNumber(int[] nums) { int num1= 0,num2 = 0; int ret = 0; for(int i = 0;i<nums.length;i++){ ret ^= nums[i]; } int pos = 0; for(;pos<32;pos++){ if((ret>>pos&1) == 1){ break; } } for(int i = 0;i<nums.length;i++){ if((nums[i]>>pos&1)==1){ num1 ^= nums[i]; }else{ num2 ^= nums[i]; } } int rets[] = {num1,num2}; return rets; } }
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LeetCode136 Single Number, LeetCode137 Single Number II, LeetCode260 Single Number III
Leetcode: SingleNumber I & II & III 136/137/260
leetcode 136:roman-to-integer&&leetcode 137:integer-to-roman