Leetcode--easy系列4

#58 Length of Last Word

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example, 
Given s = "Hello World",
return 5.

int lengthOfLastWord(char* s) {
	int count = 0;
	int len = strlen(s);
    int i = 0,j = len-1;

    while(s[j]==' ')//忽略空格
        j--;
    while(j>=i)
    {
        if(s[j] != ' ')
            count++;
        else
            break;
        j--;    
    }
	return count;
}

#66 Plus One

Given a non-negative number represented as an array of digits, plus one to the number.

The digits are stored such that the most significant digit is at the head of the list.

给定存储在字符串中的非负数，返回其+1后的结果（字符串形式）

当然字符串存储的特点是方便大数计算。所以是不能先从字符串转换为数字再+1的。

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* plusOne(int* digits, int digitsSize, int* returnSize) {
	int i = digitsSize-1;
	int *result;//返回字符串
	digits[i]++;//加1
	/*假设>=10就进位*/
	while(digits[i]>=10 && i>=1)
	{
		digits[i] = digits[i]-10;
		i--;
		digits[i]++;
	}
	/*推断最高位是否产生进位--是否添加字符串长度*/
	if(digits[0]>=10)
	{
	    *returnSize = digitsSize+1;
		result = (int *)malloc(sizeof(int)*(digitsSize+1));
		digits[0] = digits[0]-10;
		for(i=0;i<digitsSize;i++)
			result[i+1] = digits[i];
		result[0] = 1;
	}
	else
	{
	    *returnSize = digitsSize;
	    result = digits;
	}
	
	return result;
}

#67 Add Binary

Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100".

存储在字符串中的二进制数相加。

以下的代码还能够优化，边相加边进行进位推断。

測试时。a。b採用数组形式，否则假设指向字符串常量，是不容许改动字符串中的值。从而导致错误。

char* addBinary(char* a, char* b) {
    int i,j = 0,k = 0;
	int len1 = strlen(a);
	int len2 = strlen(b);
	char *p,*q,*r;
	int len_max = (len1>=len2) ?
 len1:len2;
	int len_min = (len1<=len2) ?
 len1:len2;
	
	if(len1 == 0)
		return b;
	if(len2 == 0)
		return a;
	//指针p指向 a/b中长度较长的 q指向较短的
	if(len1 >= len2)
	{
		p = a;
		q = b;
	}
	else
	{
		p = b;
		q = a;
	}
	//p=p+q---先相加----数的低位放在存储的高位
	for(j = len_min-1; j >= 0; j--)
	{
	    p[len_max-1-k] += (q[j] - '0');
	    k++;
	}
    //推断是否最高位有进位
	for(i = len_max-1; i >= 1; i--)
	{
		if(p[i] >= '2')
		{
			p[i] -= 2;
			p[i-1]++;
		}
	}
	//推断最高位
	if( p[0]-'0'<2 )
	  return p;//位数不变   
	else
	{
		p[0] -= 2;//溢出
		r = (char *)malloc(sizeof(char)*(len_max+2));
		r[0] = '1';//进位
		for(i = 1; i <= len_max; i++)
			r[i] = p[i-1];
		r[len_max+1]='\0';
		return r;
	}
}

#70 Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

类似这种问题有非常多，如猴子摘桃。走台阶，实质都是裴波那切数列。

f(n) = f(n-1)+f(n-2).

直接使用递归，例如以下。可是提交结果显示 Time Limit Exceeded

int climbStairs(int n) {
    if(n==1)
        return 1;
    if(n==2)
        return 2;
    if(n>2)
        return climbStairs(n-1)+climbStairs(n-2);
}

由于使用递归时，会反复计算。

DP算法就是保存中间结果来避免计算反复子问题。改进例如以下：

int climbStairs(int n) {
    int i,*a;
    if(n==1)
        return 1;
    if(n==2)
        return 2;
    if(n>2)
    {
        a=(int *)malloc(sizeof(int)*n);
        a[0]=1;
        a[1]=2;
        for(i=2;i<n;i++)
            a[i]=a[i-1]+a[i-2];
        return a[n-1];    
    }
}