OJ-Candy 分糖果问题

问题描述：

There are N children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

• Each child must have at least one candy.
• Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

解析：

根据等级保证高等级的比相邻的拿到糖果多，即保证为其加1即了。按着1.从前向后遍历，保证右侧的如果比左侧等级高则糖果多。 2.从后向前遍历。保证左侧比右侧的的关系。

问题延伸：

如果不是line而且ring，则需要对开始节点和结束节点进行处理。

原问题处理：

int candy(vector &ratings) {
        int size = ratings.size();
    	int *candyarray = new int[size];
    	for(int i = 0; i < size; ++i) //初始化 
    	{
    		candyarray[i] =  1;
    	}
    	for(int i = 0; i < size - 1; i++) //从前向后遍历 
    	{
    		if(ratings[i+1] > ratings[i])
    			candyarray[i+1] = candyarray[i]+1;
    	}
    	
    	for(int i = size -1; i > 0; i--) //从后向前的处理 
    	{
    		if(ratings[ i -1 ] > ratings[i] && candyarray[i -1] <= candyarray[ i ])
    		{
    			candyarray[ i - 1 ] = candyarray[i] + 1;
    		}
    	}
    
    	int sum = 0;
    	for(int i = 0 ; i < size; i++)
    	 	sum += candyarray[i];
        return sum;
    }

拓展问题的处理

 int candy(vector &ratings) {
        int size = ratings.size();
    	int *candyarray = new int[size];
    	for(int i = 0; i < size; ++i) //初始化 
    	{
    		candyarray[i] =  1;
    	}
    	for(int i = 0; i < size - 1; i++) //从前向后遍历 
    	{
    		if(ratings[i+1] > ratings[i])
    			candyarray[i+1] = candyarray[i]+1;
    	}
    	if(ratings[0] > ratings[size-1]) //对index = 0 的处理 
    	{
    		candyarray[0] = candyarray[size-1]+1;
    		for(int i = 0; i < size - 1 
    		&& ratings[i+1] > ratings[i]; i++) //!!注意不是全部遍历 
    		{
    				candyarray[i+1] = candyarray[i]+1;
    		}
    	}
    	for(int i = size -1; i > 0; i--) //从后向前的处理 
    	{
    		if(ratings[ i -1 ] > ratings[i] && candyarray[i -1] <= candyarray[ i ])
    		{
    			candyarray[ i - 1 ] = candyarray[i] + 1;
    		}
    	}
    	if(ratings[ size -1 ] > ratings[0] 
    	&& candyarray[ size - 1 ] <= candyarray[ 0 ]) //对尾部节点的处理 
    	{
    		candyarray[ size -1 ] =  candyarray[ 0 ] + 1;
    		for(int i = size -1; i > 0 && ratings[ i -1 ] > ratings[i] 
    		&& candyarray[i -1] <= candyarray[ i ]; i--)
    		{
    				candyarray[ i - 1 ] = candyarray[i] + 1;
    		}
    	}
    	int sum = 0;
    	for(int i = 0 ; i < size; i++)
    	 	sum += candyarray[i];
        return sum;
    }

感想：哈哈又好久没刷题啦~但是注意题目要求真的很重要！之所有会有拓展问题，是因为我第一次的理解有误，所以WA了。。~但是也给了自己一些提醒，比如对首末节点的处理其实不用遍历全部表。嘿嘿完成手上的翻译了，继续开始oj生活加油！！