[leet code 135]candy

1 题目

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?

2 思路

按照网上的思路，每个孩子至少有一个糖果，先从左到右遍历一遍，写出递增的糖果数，再从右到左遍历一遍完成递减的糖果数。这种大小与左右两边数据相关的问题，均可以采用这个思路。另外，有另一种空间复杂度O（1），时间复杂度O（n）的思路，可以参考http://www.cnblogs.com/felixfang/p/3620086.html。

3 代码

 public int candy(int[] ratings) {
        if(ratings == null || ratings.length == 0)
        {
            return 0;
        }
        int[] candyNums = new int[ratings.length];
        candyNums[0] = 1;  

        for(int i = 1; i < ratings.length; i++)
        {
            if(ratings[i] > ratings[i-1])  //如果第i个孩子比第i - 1孩子等级高，
            {
                candyNums[i] = candyNums[i-1]+1;
            }
            else //每人至少有一个糖果
            {
                candyNums[i] = 1;
            }
        }  

        for(int i = ratings.length-2; i >= 0; i--)
        {  

            if(ratings[i] > ratings[i + 1] && candyNums[i] <= candyNums[ i + 1]) //如果第i个孩子比第i + 1孩子等级高并且糖果比i+1糖果少
            {
                 candyNums[i] = candyNums[i + 1] + 1;
            }
        }  

        int total = 0;
        for (int i = 0; i < candyNums.length; i++) {
            total += candyNums[i];
        }

        return total;
    }