Leetcode--easy系列7

#125 Valid Palindrome

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

判断给定字符串是否是回文字符串，注意判定中忽略空格和标点（判断字符和数字），不区分字母大小写

设置2个指针i和j 进行首尾移动

//6ms
bool isPalindrome(char* s) {
    int i,j,len;
    len = strlen(s);
    if(len==0||len==1)
        return true;
    i=0;
    j=len-1;
    while(i<=j)
    {
        while(!((s[i]>='A' && s[i]<='Z') || (s[i]>='a' && s[i]<='z')) && !(s[i]>='0' && s[i]<='9'))//不是数字 或字母
           { 
               i++;
               if(i>len-1)
                    return true; 
           }
            
        while(!((s[j]>='A' && s[j]<='Z') || (s[j]>='a' && s[j]<='z')) && !(s[j]>='0' && s[j]<='9'))//不是数字 或字母
            {
                j--;
                if(j<0)
                    return false;
            }
        if(i==j)
            return true;
        if( ((s[i]>='A' && s[i]<='Z') || (s[i]>='a' && s[i]<='z'))  && ((s[j]>='A' && s[j]<='Z') || (s[j]>='a' && s[j]<='z')) )//都是字母
        {
            if(s[i]==s[j] || s[i]==s[j]+32 || s[j]==s[i]+32)//不区分大小写
            {
                i++;
                j--;
            }
            else
                return false;
        }
        else if( (s[i]>='0' && s[i]<='9') && (s[j]>='0' && s[j]<='9') )//都是数字
        {
            if(s[i]==s[j])
            {
                i++;
                j--;
            }
            else
                return false;
        }
        else
            return false;

    }
    return true;
}

一个更清晰的写法如下：

//4ms
bool isPalindrome(char* s) {
    int len = strlen(s);
    for (int i = 0, j = len - 1; i < j; i++, j--) 
    { 
        while ( (!isalnum(s[i]))  && i < j ) //isalpha(c)或isdigit(c)
            i++; 
        while ( (!isalnum(s[j])) && i < j ) 
            j--;
        if (toupper(s[i]) != toupper(s[j])) 
            return false;
    }
    return true;
}

#165 Compare Version Numbers

Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

比较版本大小：比较对应以'.'分开的数

有2种直观想法，对于两个版本数，一个方法时是根据从最前面开始取数，一个一个比较大小，只有两个数相等才继续比较下一个数，否则返回比较结果

另一种是先将这两个版本的所有数取出放在数组里，再一一对应比较大小。

下面分别给出两种方法的代码。由于Java API中存在字符串以给定字符划分的函数，给出JAVA实现。

//0ms
int compareVersion(char* version1, char* version2) {
	int i,j;
    int num1=0,num2=0;//记录以'.'分开的数字
	int start1=0,start2=0;
	int len1 = strlen(version1);
	int len2 = strlen(version2);

	while(start1<len1 && start2<len2)
	{
		num1 = 0;
		num2 = 0;
		
		for(i=start1; i<len1 && version1[i]!='.'; i++)
			num1 = (version1[i]-'0')+num1*10;
		
		for(j=start2; j<len2 && version2[j]!='.'; j++)
			num2 = (version2[j]-'0')+num2*10;

		if(num1 == num2)
		{
			start1 = i+1;//从下一个'.'开始比较下一个数字
			start2 = j+1;
		}
		else if(num1 > num2)
			return 1;
		else if(num1 < num2)
			return -1;
	}
	//特例"1.0" = "1"
	if(i==len1 && j==len2)
		return 0;
	else if(i==len1 && j<len2)
	{
		//判断j后的所有num是否为0，若为0 返回=，否则返回-1
		num2 = 0;
		while(start2<len2)
		{
			for(j=start2; j<len2 && version2[j]!='.'; j++)
				num2 = (version2[j]-'0')+num2*10;
			if(num2!=0)
				return -1;
			num2 = 0;//置0，下一次判断
			start2++;
		}
		return 0;
	}
		
	else if(i<len1 && j==len2)
	{
			//判断j后的所有num是否为0，若为0 返回=，否则返回-1
		num1 = 0;
		while(start1 < len1)
		{
			for(i=start1; i<len1 && version1[i]!='.'; i++)
				num1 = (version1[i]-'0')+num1*10;
			if(num1!=0)
				return 1;
			num1 = 0;//置0，下一次判断
			start1++;
		}
		return 0;
	}
}

//0ms
int compareVersion(char* version1, char* version2) {
	int nums1[200]={0},nums2[200]={0};//先假设版本号最多有5个
	int start1=0,start2=0;//指向每一个char
	int len1 = strlen(version1);
	int len2 = strlen(version2);
	int i;
	int size1=0,size2=0;
	int temp=0;//数字
	while(start1 < len1)
	{
		for(i=start1; i<len1 && version1[i]!='.'; i++)
			temp = (version1[i]-'0')+temp*10;
		if(i!=start1)//!!!
			nums1[size1++] = temp;
		temp = 0;//置0，下一次计算
		start1 = i+1;;
	}
	while(start2 < len2)
	{
		for(i=start2; i<len2 && version2[i]!='.'; i++)
			temp = (version2[i]-'0')+temp*10;
		if(i!=start2)
			nums2[size2++] = temp;
		temp = 0;//置0，下一次计算
		start2 = i+1;
	}
	//比较nums1[]和nums2[]的每一个元素
//	temp = (size1>size2)?(size1-1):(size2-1);
	for(i=0; i<200;i++)
	{
		if(nums1[i] > nums2[i])
			return 1;
		else if(nums1[i] < nums2[i])
			return -1;
	}
	return 0;
}

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");

        int longest = v1.length > v2.length? v1.length: v2.length;

        for(int i=0; i<longest; i++)
        {
            int ver1 = i<v1.length? Integer.parseInt(v1[i]): 0;
            int ver2 = i<v2.length? Integer.parseInt(v2[i]): 0;

            if(ver1> ver2) return 1;
            if(ver1 < ver2) return -1;
        }
        return 0;
    }
}

#169 majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

n个元素的数组，某个元素至少重复  ⌊ n/2 ⌋ + 1  个 ，找出该元素。

任意两个不同的元素相消，最后剩下的即为该元素。也可以先排序，再取中间元素

//8ms

int majorityElement(int* nums, int numsSize) {
    int major = nums[0];//初值
    int count = 1;
    int i;
    for(i = 1; i<numsSize; i++)
    {
        if(nums[i]==major)//如果相同，计数+1
            count++;
        else
        {
            count--;//如果不同，计数-1，直到为0，则重置新的major
            if(count==0)
            {
                major = nums[i];
                count = 1;
            }
        }
    }
    return major;
}

#171 Excel Sheet Column Number

Given a column title as appear in an Excel sheet, return its corresponding column number.

For example:

    A -> 1
    B -> 2
    C -> 3
    ...
    Z -> 26
    AA -> 27
    AB -> 28 

类似进制转换

//4ms
int titleToNumber(char* s) {
    int len,i,j=0,num=0;
	len = strlen(s);

	for(i=len-1;i>=0;i--)
	{
		num += (s[i]-'A'+1)*pow(26,j);
		j++;
	}
	return num;
}