[LeetCode] Candy
2014-08-19 20:08
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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?
思路:
从左往右扫描权重数组,保证权重大于左邻居的权重时得到的糖果也比座邻居大多,可以得到一个糖果数量数组。
从右往左扫描,满足右边邻居条件,得到另外一个糖果数量数组。
取每个孩子在两个糖果数量数组大的为最终值
public class Solution {
public int candy(int[] ratings) {
if (ratings == null) {
return 0;
}
int len = ratings.length;
if (len < 1) {
return 0;
}
int[] leftArr = new int[len];
int[] rightArr = new int[len];
leftArr[0] = 1;
for (int i = 1; i < len; i++) {
if (ratings[i] > ratings[i - 1]) {
leftArr[i] = leftArr[i - 1] + 1;
} else {
leftArr[i] = 1;
}
}
rightArr[len - 1] = 1;
for (int i = len - 2; i >= 0; i--) {
if (ratings[i] > ratings[i + 1]) {
rightArr[i] = rightArr[i + 1] + 1;
} else {
rightArr[i] = 1;
}
}
int total = 0;
for (int i = 0; i < len; i++) {
total += (leftArr[i] > rightArr[i] ? leftArr[i] : rightArr[i]);
}
return total;
}
}
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:
从左往右扫描权重数组,保证权重大于左邻居的权重时得到的糖果也比座邻居大多,可以得到一个糖果数量数组。
从右往左扫描,满足右边邻居条件,得到另外一个糖果数量数组。
取每个孩子在两个糖果数量数组大的为最终值
public class Solution {
public int candy(int[] ratings) {
if (ratings == null) {
return 0;
}
int len = ratings.length;
if (len < 1) {
return 0;
}
int[] leftArr = new int[len];
int[] rightArr = new int[len];
leftArr[0] = 1;
for (int i = 1; i < len; i++) {
if (ratings[i] > ratings[i - 1]) {
leftArr[i] = leftArr[i - 1] + 1;
} else {
leftArr[i] = 1;
}
}
rightArr[len - 1] = 1;
for (int i = len - 2; i >= 0; i--) {
if (ratings[i] > ratings[i + 1]) {
rightArr[i] = rightArr[i + 1] + 1;
} else {
rightArr[i] = 1;
}
}
int total = 0;
for (int i = 0; i < len; i++) {
total += (leftArr[i] > rightArr[i] ? leftArr[i] : rightArr[i]);
}
return total;
}
}
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