标签：des style blog io ar color os sp for
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:
What is the minimum candies you must give?
注意，像 6 5 5 4 这样的 可以分 2 1 2 1 就是数字相同的不需要糖果数相同
我的思路，分别用两个vector存储递增和递减所需要的糖果，最终结果取两个的最大值
如 权重 6 6 1 5 4 4 3 7 4 2 2 2 5 4 3 5
找递增 1 1 1 2 1 1 1 2 1 1 1 1 2 1 1 2 从前向后找，初始都是1，遇到递增的就加上他前面查找的数字
找递减 1 2 1 2 1 2 1 3 2 1 1 1 3 2 1 1 从后向前找，初始都是1，遇到递增的就加上他前面查找的数字
最终 1 2 1 2 1 2 1 3 2 1 1 1 3 2 1 2
int candy(vector<int> &ratings) { int ans = 0; vector<int> candysUp(ratings.size(), 1); vector<int> candysDown(ratings.size(), 1); for(int i = 1; i < ratings.size(); i++) //查找递增 { if(ratings[i] > ratings[i - 1]) { candysUp[i] += candysUp[i - 1]; } } for(int i = ratings.size() - 2; i >= 0; i--) //查找递减 { if(ratings[i + 1] < ratings[i]) { candysDown[i] += candysDown[i + 1]; } } for(int i = 0; i < ratings.size(); i++) { ans += max(candysUp[i], candysDown[i]); } return ans; }
大神只循环一遍的思路：
其实整体思想跟上面一样，但是上面需要循环很多次是因为递增和递减分别处理。因为递减时不知道最大的数应该加几。
大神的思路是递增的就像上面那样加，但是递减的，先都加1，如果后面仍是递减的，再把之前少加的给加回来（如前面有nSeqLen个数字少加了1，就把答案直接多加一个nSeqLen）。
//大神循环一遍的代码 int candy2(vector<int> &ratings) { int nCandyCnt = 0;///Total candies int nSeqLen = 0; /// Continuous ratings descending sequence length int nPreCanCnt = 1; /// Previous child‘s candy count int nMaxCntInSeq = nPreCanCnt; if(ratings.begin() != ratings.end()) { nCandyCnt++;//Counting the first child‘s candy. for(vector<int>::iterator i = ratings.begin()+1; i!= ratings.end(); i++) { // if r[k]>r[k+1]>r[k+2]...>r[k+n],r[k+n]<=r[k+n+1], // r[i] needs n-(i-k)+(Pre‘s) candies(k<i<k+n) // But if possible, we can allocate one candy to the child, // and with the sequence extends, add the child‘s candy by one // until the child‘s candy reaches that of the prev‘s. // Then increase the pre‘s candy as well. // if r[k] < r[k+1], r[k+1] needs one more candy than r[k] if(*i < *(i-1)) { //Now we are in a sequence nSeqLen++; if(nMaxCntInSeq == nSeqLen) { //The first child in the sequence has the same candy as the prev //The prev should be included in the sequence. nSeqLen++; } nCandyCnt+= nSeqLen; nPreCanCnt = 1; } else { if(*i > *(i-1)) { nPreCanCnt++; } else { nPreCanCnt = 1; } nCandyCnt += nPreCanCnt; nSeqLen = 0; nMaxCntInSeq = nPreCanCnt; } } } return nCandyCnt; }
【leetcode】Candy（hard) 自己做出来了 但别人的更好
标签：des style blog io ar color os sp for
原文地址：http://www.cnblogs.com/dplearning/p/4134923.html
