leetcode 135 —— 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?

思路1：用栈保存递减序列（估计只有我看得懂。。。）

class Solution {
public:
int candy(vector<int>& ratings) {
int n = ratings.size();
if (n == 0) return 0;
vector<int> s(n, 0);
stack<int> stk;
int i = 0;
s[i] = 1;
stk.push(i);
i++;
while (i < n){
if (ratings[i - 1] < ratings[i]){ //开始变大
stk.pop();
if (!stk.empty()||i==1){
s[i - 1] = 1;
s[i] = 2;
}
else
s[i] = s[i - 1] + 1;
while (!stk.empty()){
int t = stk.top();
stk.pop();
if (ratings[t] == ratings[t + 1])
s[t] = 1;
else
s[t] = s[t + 1] + 1;
if (stk.empty()&&t>0)
s[t] = max(s[t], s[t - 1] + 1);
}
stk.push(i);
}
else
stk.push(i);
i++;
}
if (!stk.empty()){
int t = stk.top();
stk.pop();
if (!stk.empty())
s[t] = 1;
else
s[t] = s[t - 1] + 1;
}
while (!stk.empty()){
int t = stk.top();
s[t] = ratings[t] == ratings[t + 1] ? 1 : s[t + 1] + 1;
stk.pop();
if (stk.empty()&&t>0)
s[t] = max(s[t], s[t - 1] + 1);
}
int res = 0;
for (int i = 0; i < n; i++)
res += s[i];

return res;
}
};
思路2：从左往右扫描一遍，使右边大的比左边拿得多

    从右往左扫描一遍，使左边大的比右边拿得多

  代码更加简洁了

class Solution {
public:
int candy(vector<int>& ratings) {
int n = ratings.size();
if (n == 0) return -1;
vector<int> s(n, 1);
for (int i = 0; i < n-1; i++){
if (ratings[i] < ratings[i + 1])
s[i + 1] = s[i] + 1;
}
for (int i = n - 1; i >= 1; i--){
if (ratings[i - 1]>ratings[i]&&s[i-1]<=s[i]){
s[i - 1] = s[i] + 1;
}
}
int res = 0;
for (int i = 0; i < n; i++)
res += s[i];
return res;
}
};