2014
11-18

# LeetCode-Candy[数组]

### 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?

// LeetCode, Candy
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
int candy(vector<int> &ratings) {
const int n = ratings.size();
vector<int> increment(n);

// 左右各扫描一遍
for (int i = 1, inc = 1; i < n; i++) {
if (ratings[i] > ratings[i - 1])
increment[i] = max(inc++, increment[i]);
else
inc = 1;
}

for (int i = n - 2, inc = 1; i >= 0; i--) {
if (ratings[i] > ratings[i + 1])
increment[i] = max(inc++, increment[i]);
else
inc = 1;
}
// 初始值为n，因为每个小朋友至少一颗糖
return accumulate(&increment[0], &increment[0]+n, n);
}
};


// LeetCode, Candy
// 备忘录法，时间复杂度O(n)，空间复杂度O(n)
// @author fancymouse (http://weibo.com/u/1928162822)
class Solution {
public:
int candy(const vector<int>& ratings) {
vector<int> f(ratings.size());
int sum = 0;
for (int i = 0; i < ratings.size(); ++i)
sum += solve(ratings, f, i);
return sum;
}
int solve(const vector<int>& ratings, vector<int>& f, int i) {
if (f[i] == 0) {
f[i] = 1;
if (i > 0 && ratings[i] > ratings[i - 1])
f[i] = max(f[i], solve(ratings, f, i - 1) + 1);
if (i < ratings.size() - 1 && ratings[i] > ratings[i + 1])
f[i] = max(f[i], solve(ratings, f, i + 1) + 1);
}
return f[i];
}
};


Java代码:

public class Solution {
public int candy(int[] ratings) {
int cntArr[] = new int[ratings.length];
for(int i=0; i<cntArr.length; i++) cntArr[i]=1;
for(int i=1; i<cntArr.length; i++){
if(ratings[i] > ratings[i-1])
cntArr[i] = cntArr[i-1]+1;
}
int ans = cntArr[cntArr.length-1];
for(int i=cntArr.length-2; i>=0; i--){
if(ratings[i] > ratings[i+1] && cntArr[i] <= cntArr[i+1])
cntArr[i] = cntArr[i+1]+1;
ans += cntArr[i];
}
return ans;

}
}

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