LeetCode OJ

感觉这道题很有趣。一开始的想法没有经过深入的思考，总是想当然，自然也就容易出错。后来深入思考题中提出的两个条件，想到一个局部极小这个概念，接下来的想法，就显得很自然而然。

总的来说是先DP，然后根据DP的结果决定给每个小孩多少个candies。DP的关键在于定义一个表示子问题的数组变量，在这个题中定义两个这样的变量：

left[i]表示在i的左边，小于ratings[i]，且从最近的一个极小值开始连续到i-1（包括i-1），的个数；

right[i]表示在i的右边，小于ratings[i]，且从最近的一个极小值开始连续到i+1（包括i+1），的个数；

之后就是决策过程，我觉得把图画出来，一切尽在图中了：

下面是AC代码：

/**
     * 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?
     * @param ratings
     * @return
     */
    public int candy(int[] ratings)
    {
        int N = ratings.length;
        int[] candy = new int [N]; //who get how many candies
        int[] left = new int[N]; //left[i] presents the number of consecutive ratings (in left side of i) smaller than ratings[i]
        int[] right = new int[N]; //right[i] presents the number of consecutive ratings (in right side of i) smaller than ratings[i] 
        
        //by DP, caculating the left and right
        left[0] = 0;
        for(int i=1;i<N;i++)
            left[i] = ratings[i]>ratings[i-1]? left[i-1]+1:0;
        right[N-1] = 0;
        for(int i=N-2;i>=0;i--)
            right[i] = ratings[i]>ratings[i+1]? right[i+1]+1:0;
        //decide how many candies each one should get
        int r = 0; 
        for(int i=0;i<N;i++)
        {
            //the local minimum
            if(left[i] == 0 && right[i] == 0)
                candy[i] = 1;
            //the local maximum
            if(left[i]!=0 && right[i]!=0)
                candy[i] = Math.max(left[i], right[i])+1;
            //other cases
            else 
                candy[i] = left[i]!=0? left[i]+1:right[i]+1;
            r =r+candy[i];
        }
        return r;
    }