lintcode-分糖果

空间复杂度O(n)

class Solution {
public:
    /**
     * @param ratings Children's ratings
     * @return the minimum candies you must give
     */
    int candy(vector& ratings) {
        // Write your code here
        
        vector dp(ratings.size(), 1);
        
        for(int i = 1; i < dp.size(); ++i){
            if(ratings[i] > ratings[i-1]){
                dp[i] = max(dp[i], dp[i-1] + 1);
            }
        }
        
        for(int i = dp.size() - 1; i > 0; --i){
            if(ratings[i] < ratings[i-1]){
                dp[i-1] = max(dp[i-1], dp[i] + 1);
            }
        }
        
        return accumulate(dp.begin(), dp.end(), 0);
    }
};

空间复杂度O(1)

class Solution {
public:
    /**
     * @param ratings Children's ratings
     * @return the minimum candies you must give
     */
    int candy(vector& ratings) {
        // Write your code here
        
        int total = 0;
        int preCount = 1;  //前一个小孩的糖果数
        int descendStartCount = preCount;  //递减序列开始的孩子的糖果数
        int length = 0;  
        
        if(ratings.size() == 0) {
            return 0;
        }
        
        total++;
        for(int i = 1; i < ratings.size(); ++i) {
            if(ratings[i] < ratings[i-1]) {
                length ++ ;
                if(descendStartCount <= length) {
                    total++;  //根据递减序列的长度修正递减序列第一个孩子的糖果数
                }
                total += length;  //逆序获取后续孩子的糖果数
                preCount = 1;     //注意
            }else{
                int currentCount = 0;
                if(ratings[i] > ratings[i-1]) {
                    currentCount = preCount + 1;
                }else{
                    currentCount = 1;
                }
                
                total += currentCount;
                preCount = currentCount;
                length = 0;      //注意
                descendStartCount = currentCount;
            }
        }
        
        return total;
        
    }
};