leetcode135. Candy
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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?
问题如上。
先放上一种方法的实现,以下代码在不考虑数组越界的情况下可以输出正确结果。
1 class Solution{ 2 public: 3 int candy(vector<int> &ratings){ 4 int start = 0, end, count = 0; //start=right 5 int candyCount[1000]; //储存每个人的糖果数量 6 ratings.push_back(-1); 7 if (ratings.size() == 2) return 1; 8 for (int i = 0; i < ratings.size(); ++i){ 9 if ((ratings[i] <= ratings[i - 1] && ratings[i] < ratings[i + 1]) || (i - 1 < 0 && ratings[i] < ratings[i + 1]) || (i + 1 > ratings.size() - 1 && ratings[i] < ratings[i - 1])){ 10 candyCount[i] = 1; 11 end = i; 12 if (end == start) continue; 13 vector<int>::iterator maxelem = max_element(ratings.begin() + start, ratings.begin() + end); 14 int flag = std::distance(ratings.begin(), maxelem); 15 for (int m = start + 1; m < flag; ++m) 16 ratings[m] > ratings[m - 1] ? candyCount[m] = candyCount[m - 1] + 1 : candyCount[m] = candyCount[m - 1]; 17 for (int n = end - 1; n > flag; --n){ 18 ratings[n] > ratings[n + 1] ? candyCount[n] = candyCount[n + 1] + 1 : candyCount[n] = candyCount[n + 1]; 19 if ((end == ratings.size() - 1) && (n == end - 1)) candyCount[n]--; 20 } 21 if (ratings[flag] == ratings[flag + 1]){ 22 candyCount[flag - 1] < candyCount[flag + 1] ? candyCount[flag] = candyCount[flag + 1] : candyCount[flag] = candyCount[flag - 1] + 1; 23 while ((ratings[flag] == ratings[flag + 1])&&(flag+1<ratings.size())) 24 candyCount[flag + 1] = candyCount[flag++]; 25 }else{ 26 candyCount[flag - 1] > candyCount[flag + 1] ? candyCount[flag] = candyCount[flag - 1] + 1 : candyCount[flag] = candyCount[flag + 1] + 1; 27 } 28 start = i; 29 } 30 } 31 for (int k = 0; k < ratings.size() - 1; ++k) 32 count += candyCount[k]; 33 return count; 34 } 35 };
写的又臭又长。。。
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Baozi Leetcode solution 135: Candy