LeetCode力扣之135. Candy
2018-03-27 10:17
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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?package leetCode;
import java.util.Arrays;
/**
* Created by lxw, liwei4939@126.com on 2018/3/27.
*/
public class L135_Candy {
/*从左到右遍历一次,再从右到左遍历一次,具体地,
* 从左到右遍历,使得右边评分更高的元素比左边至少多于1个糖果
* 从右到左遍历,使得左边评分更高的元素比右边至少多于1个糖果
* */
public int candy(int[] ratings){
int[] candies = new int[ratings.length];
Arrays.fill(candies, 1);
for (int i = 1; i < candies.length; i++){
if (ratings[i] > ratings[i-1]){
candies[i] = candies[i-1] + 1;
}
}
for (int i = candies.length-2; i >= 0; i--){
if (ratings[i] > ratings[i+1]){
candies[i] = Math.max(candies[i],candies[i+1] +1);
}
}
int sum = 0;
for (int candy : candies){
sum += candy;
}
return sum;
}
}
Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?package leetCode;
import java.util.Arrays;
/**
* Created by lxw, liwei4939@126.com on 2018/3/27.
*/
public class L135_Candy {
/*从左到右遍历一次,再从右到左遍历一次,具体地,
* 从左到右遍历,使得右边评分更高的元素比左边至少多于1个糖果
* 从右到左遍历,使得左边评分更高的元素比右边至少多于1个糖果
* */
public int candy(int[] ratings){
int[] candies = new int[ratings.length];
Arrays.fill(candies, 1);
for (int i = 1; i < candies.length; i++){
if (ratings[i] > ratings[i-1]){
candies[i] = candies[i-1] + 1;
}
}
for (int i = candies.length-2; i >= 0; i--){
if (ratings[i] > ratings[i+1]){
candies[i] = Math.max(candies[i],candies[i+1] +1);
}
}
int sum = 0;
for (int candy : candies){
sum += candy;
}
return sum;
}
}
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