1. single-number

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?

给定一个数组,每个元素都出现2次除了其中的一个,找出只出现一次的数字注意:算法必须是线性时间复杂度,可以不使用额外空间实现吗?

//异或运算,不同为1 相同为0
public int singleNumber(int[] A) {
int x = 0;
for (int a : A) {
x = x ^ a;
}
return x;
}

2.single-number-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?

除了一个元素外,其它元素都是出现三次,求那个元素?

如果是除了一个元素,其它的都出现两次,则所有的数异或就是结果。

int 数据共有32位,可以用32变量存储 这 N 个元素中各个二进制位上  1  出现的次数,最后 在进行 模三 操作,如果为1,那说明这一位是要找元素二进制表示中为 1 的那一位。

public class Solution{
public int singleNumber(int [] A){
if(A==null || A.length ==0){
return -1;
}
int result=0;
int[] bits=new int[32];
for(int i=0;i<32;i++){
for(int j=0;j<A.length;j++){
bits[i]+=A[j]>>i&1;
bits[i]%=3;
}
result |=(bits[i] << i);
}
return result;
}
}

3.reverse-integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

public int reverse(int x) {
int rev = 0;
while(x != 0){
rev = rev*10 + x%10;
x = x/10;
} return rev;
}
 
public int reverse(int x) {
//flag marks if x is negative
boolean flag = false;
if (x < 0) {
x = 0 - x;
flag = true;
} int res = 0;
int p = x; while (p > 0) {
int mod = p % 10;
p = p / 10;
res = res * 10 + mod;
} if (flag) {
res = 0 - res;
} return res;
}

4.longest-common-prefix

Write a function to find the longest common prefix string amongst an array of strings.

找出所有字符串的最长公共前缀

public class Solution {

    // 1. Method 1, start from the first one, compare prefix with next string, until end;
// 2. Method 2, start from the first char, compare it with all string, and then the second char
// I am using method 1 here
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
}
String prefix = strs[0];
for(int i = 1; i < strs.length; i++) {
int j = 0;
while( j < strs[i].length() && j < prefix.length() && strs[i].charAt(j) == prefix.charAt(j)) {
j++;
}
if( j == 0) {
return "";
}
prefix = prefix.substring(0, j);
}
return prefix;
} }

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