Leetcode--easy系列3

#26 Remove Duplicates from Sorted Array

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

------这个题提交了好几次才AC，主要是函数不仅要返回不反复数的个数count。还要求将不反复的数放置在原始数组的前count个位置。阅读理解非常重要啊。

int removeDuplicates(int* nums, int numsSize) {
	int count=1;//14ms
	int i;
	if(numsSize==0)
		return 0;
	if(numsSize==1)
		return 1;
	for(i=1;i<=numsSize-1;i++)
	{
		if(nums[i]!=nums[i-1])
		{
			nums[count] = nums[i];//不仅要求出不反复个数，还要将不反复元素放在前面
			count++;
		}
	}
	return count;
}

#27 Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

从数组中去除指定元素。返回剩余元素个数count。-------同一时候，原来数组中前count个元素即为原来数组的剩余元素。否则不会AC。

写了2种方法，法一时间复杂度和空间复杂度都比較高，但便于理解。法二使用了双指针，i和count分别指向原来的元素和非指定删除值的元素

//0ms
int removeElement(int* nums, int numsSize, int val) {
    <span style="white-space:pre">	</span>int i,j = 0,count = 0;
	int *a;
	a = (int *)malloc(sizeof(int)*numsSize);
	for(i=0; i < numsSize; i++)
	{
		if(nums[i] == val)
			count++;
		else
			a[j++]=nums[i];
			
	}
	for(i=0; i < j; i++)
		nums[i] = a[i];
	return numsSize-count;
}

//0ms
int removeElement(int* nums, int numsSize, int val) {
    <span style="white-space:pre">	</span>int i = 0,count = 0;
<span style="white-space:pre">	</span>while(i < numsSize)
	{
		if(nums[i] == val)
			i++;
		else 
			nums[count++] = nums[i++];
	}
	return count;
}

#28 Implement strStr()

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

经典的串的模式匹配问题，主要有BF和KMP算法，详细解析可见本博客相关博客《串模式匹配之BF和KMP算法》

//BF
int strStr(char* haystack, char* needle) {
    int i=0,j=0,k;
	int len1 = strlen(haystack);
	int len2 = strlen(needle);
	if(len2==0)
	    return 0;
	if(len1==0&&len2!=0)
	    return -1;
	    
	while( i<len1 && j<len2)
	{
		if(haystack[i]==needle[j])
		{
			i++;
			j++;
		}
		else
		{
			i=i-j+1;
			j=0;
		}
	}
	if(j>=len2)
		k=i-len2;
	else
		k=-1;
	return k;
}

#36 Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

推断给定数独是否是一个有效数独：仅仅考虑没有反复数字。最直观的解法是推断每一列，每一行。每个3*3的方块内不包括反复的数字

直接贴代码

bool isValidSudoku(char** board, int boardRowSize, int boardColSize) {
        int hash[10];
	int i,j,k,m,n;
	char small[3][3];
	memset(hash,0,sizeof(hash));
	if(boardRowSize%3 != 0 || boardColSize%3 != 0)
		return false;
	for(i=0; i < boardRowSize; i++)
		for(j=0;j<boardColSize;j++)
		{
			if(board[i][j] == '.'||(board[i][j] <= '9' && board[i][j] >= '1'))
				continue;
			else
				return false;
		}
	for(i = 0; i < boardRowSize; i++)
	{
	    memset(hash,0,sizeof(hash));
		for(j = 0; j < boardColSize; j++)
			if(board[i][j] != '.')
				hash[board[i][j] - '0']++;

		for(k = 1;k < 10;k++)
			if(hash[k] > 1)
				return false;
	}	
	for(j = 0; j < boardColSize; j++)
	{
	    memset(hash,0,sizeof(hash));
		for(i = 0; i < boardRowSize; i++)
			if(board[i][j] != '.')
				hash[board[i][j] - '0']++;

		for(k = 1; k < 10; k++)
			if(hash[k] > 1)
				return false;
	}
	memset(hash,0,sizeof(hash));

	for(i = 0; i < boardRowSize; i = i+3)
		for(j = 0; j < boardColSize; j = j+3)
		{

			small[0][0] = board[i][j];
			small[0][1] = board[i][j+1];
			small[0][2] = board[i][j+2];
			small[1][0] = board[i+1][j];
			small[1][1] = board[i+1][j+1];
			small[1][2] = board[i+1][j+2];
			small[2][0] = board[i+2][j];
			small[2][1] = board[i+2][j+1];
			small[2][2] = board[i+2][j+2];
			for(m=0; m < 3; m++)
				for(n = 0; n < 3; n++)
				{
					if(small[m][n] != '.')
						hash[small[m][n] - '0']++;
				}

		for(k=1; k < 10; k++)
			if(hash[k] > 1)
				return false;
		memset(hash,0,sizeof(hash));		
		}
	return true;
}

简化后例如以下：

bool isParticallyValid(char** board,int x1,int y1,int x2,int y2)
{
	int hash[10],i,j;
	memset(hash,0,sizeof(hash));
	for(i = x1; i <= x2; i++)
	{
		for(j = y1; j <= y2; j++)
		{
			if(board[i][j] != '.')
			{
			    hash[board[i][j]-'0']++;
			    if(hash[board[i][j]-'0'] > 1)
			        return false;
			}
		}
	}
	return true;
}
bool isValidSudoku(char** board, int boardRowSize, int boardColSize) 
{
	int i,j;
	//判定每一行每一列是否包括反复元素
	for(i = 0; i < 9; i++)
	{
		if(!isParticallyValid(board,i,0,i,8))
			return false;
		if(!isParticallyValid(board,0,i,8,i))
			return false;
	}
	//判定3*3的方块内是否包括反复元素
	for(i = 0; i < 3; i++)
	{
		for(j = 0; j < 3; j++)
		{
			if(!isParticallyValid(board,i*3,j*3,i*3+2,j*3+2))
				return false;
		}
	}
	return true;
}