[LeetCode]Candy

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?

Analysis:

the value of candy[i] depends on its left and right node. the idea comes from here, we can scan from two sides

Scan the rating array from left to right and then from right to left. In every scan just consider the rising order (l->r:
r[i]>r[i-1] or r->l: r[i]>r[i+1]), assign +1 candies to the rising position.

The final candy array is the maximum (max(right[i],left[i])) in each position.

The total candies is the sum of the final candy array.



java

public int candy(int[] ratings) {
		int len = ratings.length;
		if(len<=0) return 0;
		if(len==1) return 1;
		int result = 0;
		int []lr = new int[len];
		int []rl = new int[len];
		lr[0] = 1;
		rl[len-1] = 1;
		for(int i=1;i<len;i++){//from left to right
			if(ratings[i]>ratings[i-1])
				lr[i] = lr[i-1]+1;
			else {
				lr[i] = 1;
			}
		}
		for(int i=len-2;i>=0;i--){
			if(ratings[i]>ratings[i+1])
				rl[i] = rl[i+1]+1;
			else
				rl[i] = 1;
		}
		for(int i=0;i<len;i++){
			result+=Math.max(lr[i], rl[i]);
		}
		return result;
    }

c++

int candy(vector<int> &ratings) {
       int result =0;
    vector<int> lc(ratings.size(),1);
    vector<int> rc(ratings.size(),1);
    for(int i=1;i<ratings.size();i++){
        if(ratings[i]>ratings[i-1])
            lc[i] = lc[i-1]+1;
    }
    for(int i=ratings.size()-2;i>=0;i--){
        if(ratings[i]>ratings[i+1])
            rc[i] = rc[i+1]+1;
    }
    for(int i=0;i<ratings.size();i++){
        result+=max(lc[i],rc[i]);
    }
    return result;
    }