Candy leetcode java
2014-07-30 08:50
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题目:
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?
题解:
这道题和Trapping water那个是一样的想法,因为无论是水坑还是得到糖的小朋友,影响因素都不只一边,都是左右两边的最小值/最大值来决定的。
所以这道题跟上一道一样,也是左右两边遍历数组。
leftnums数组存从左边遍历,当前小朋友对比其左边小朋友,他能拿到糖的数量;
rightnums数组存从右边遍历,当前小朋友对比其右边小朋友,他能拿到的糖的数量。
最后针对这两个数组,每个小朋友能拿到的糖的数量就是这两个数最大的那个数,求总加和就好了。
代码如下:
1 public int candy(int[] ratings) {
2 if(ratings==null || ratings.length==0)
3 return 0;
4
5 int[] leftnums = new int[ratings.length];
6 int[] rightnums = new int[ratings.length];
7
8 leftnums[0]=1;
9 for(int i=1;i<ratings.length;i++){
if(ratings[i]>ratings[i-1])
leftnums[i] = leftnums[i-1]+1;
else
leftnums[i] = 1;
}
rightnums[ratings.length-1] = leftnums[ratings.length-1];
for(int i=ratings.length-2;i>=0;i--){
if(ratings[i]>ratings[i+1])
rightnums[i] = rightnums[i+1]+1;
else
rightnums[i] = 1;
}
int res = 0;
for(int i = 0; i<ratings.length; i++)
res += Math.max(leftnums[i],rightnums[i]);
return res;
}
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?
题解:
这道题和Trapping water那个是一样的想法,因为无论是水坑还是得到糖的小朋友,影响因素都不只一边,都是左右两边的最小值/最大值来决定的。
所以这道题跟上一道一样,也是左右两边遍历数组。
leftnums数组存从左边遍历,当前小朋友对比其左边小朋友,他能拿到糖的数量;
rightnums数组存从右边遍历,当前小朋友对比其右边小朋友,他能拿到的糖的数量。
最后针对这两个数组,每个小朋友能拿到的糖的数量就是这两个数最大的那个数,求总加和就好了。
代码如下:
1 public int candy(int[] ratings) {
2 if(ratings==null || ratings.length==0)
3 return 0;
4
5 int[] leftnums = new int[ratings.length];
6 int[] rightnums = new int[ratings.length];
7
8 leftnums[0]=1;
9 for(int i=1;i<ratings.length;i++){
if(ratings[i]>ratings[i-1])
leftnums[i] = leftnums[i-1]+1;
else
leftnums[i] = 1;
}
rightnums[ratings.length-1] = leftnums[ratings.length-1];
for(int i=ratings.length-2;i>=0;i--){
if(ratings[i]>ratings[i+1])
rightnums[i] = rightnums[i+1]+1;
else
rightnums[i] = 1;
}
int res = 0;
for(int i = 0; i<ratings.length; i++)
res += Math.max(leftnums[i],rightnums[i]);
return res;
}
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