LeetCode--137. Single Number II && 260. Single Number III之暴力搜法
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137. Single Number II
题目:
Given an 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?
分析:给一个整数数组,其中除了一个数外其他整数的数目均为三个,找出单个的那个整数,本代码使用暴力搜索法。分情况处理,排序后单个数在最后或在中间和开头的结构不一样,分情况处理。
代码:
public class Solution {
public int singleNumber(int[] nums) {
if(nums.length==2||nums.length==3) return 0;
if(nums.length==1){//长度为1,直接输出第一个数即可
return nums[0];
}else{
Arrays.sort(nums);//对输入的数排序
if(nums[nums.length-2]!= nums[nums.length-1]){//单个数在最后,直接输出最后一个数即可
return nums[nums.length-1];
}else{
int result=0;
for(int i=0,j=1,k=2;i<nums.length-1;){//单个数在中间和头部的情况
if((nums[i] == nums[j])&&(nums[j]==nums[k])){
i=i+3;
j=j+3;
k=k+3;
}else{
result = nums[i];
break;
}
}
return result;
}
}
}
}
260. Single Number III
题目:
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
分析:给一个整数数组,除了两个整数外,其他的整数的数目均为两个,找到两个单个的整数,分情况处理,排序后两个单个的整数在头部,尾部,中间,一中间一尾部处理
代码:
public class Solution {
public int[] singleNumber(int[] nums) {
int result[] = new int[2];
if(nums.length==2){//长度为2,直接输出即可
return nums;
}else{
Arrays.sort(nums);//对输入的数排序
if((nums[nums.length-2]!= nums[nums.length-1])&&(nums[nums.length-2]!= nums[nums.length-3])){//排序后单个数在最后两个,直接输出最后两数即可
result[0]=nums[nums.length-2];
result[1]=nums[nums.length-1];
return result;
}else if((nums[0]!= nums[1])&&(nums[1]!= nums[2])){//排序后两个单数在首位的情况
result[0]=nums[0];
result[1]=nums[1];
return result;
}else if((nums[nums.length-2]!= nums[nums.length-1])&&(nums[nums.length-2]==nums[nums.length-3])){//一个在尾部的情况
result[1] = nums[nums.length-1];
for(int i=0,j=1;i<nums.length-1;){//排序后另外一个单数在头部或中间的情况
if(nums[i] == nums[j]){
i=i+2;
j=j+2;}
else{
result[0]=nums[i];
break;
}
}
return result;
}else{
ArrayList<Integer> allList = new ArrayList<Integer>();
for(int i=0,j=1;i<nums.length-1;){//排序后两个单数在中间的情况
if(nums[i] == nums[j]){
i=i+2;
j=j+2;
}else{
allList.add(nums[i]);
i=i+1;
j=j+1;
}
}
Integer tmpInteger[] = new Integer[2];
allList.toArray(tmpInteger);
int ary[] = new int[2];
for (int i = 0; i < 2; i++) {
ary[i] = tmpInteger[i].intValue();
}
return ary;
}
}
}
}
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