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?
解题之法
class Solution {
public:
int candy(vector<int>& ratings) {
int res = 0;
vector<int> num(ratings.size(), 1);
for (int i = 0; i < (int)ratings.size() - 1; ++i) {
if (ratings[i + 1] > ratings[i]) num[i + 1] = num[i] + 1;
}
for (int i = (int)ratings.size() - 1; i > 0; --i) {
if (ratings[i - 1] > ratings[i]) num[i - 1] = max(num[i] + 1, num[i - 1]);
}
for (int i = 0; i < num.size(); ++i) {
res += num[i];
}
return res;
}
};
分析
这道题看起来很难，其实解法并没有那么复杂。
首先初始化每个人一个糖果，然后这个算法需要遍历两遍，第一遍从左向右遍历，如果右边的小盆友的等级高，等加一个糖果，这样保证了一个方向上高等级的糖果多。
然后再从右向左遍历一遍，如果相邻两个左边的等级高，而左边的糖果又少的话，则左边糖果数为右边糖果数加一。
最后再把所有小盆友的糖果数都加起来返回即可。
