leetcode: 复杂度

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;
    }
 
}