leetcode 135. 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?

讲的是一列小朋友按排名值分糖果

翻译过来就是：一个数组，每个元素最少赋值1，排名值高的元素要比相邻元素赋值高

这个题目我认为最关键的是建立起模型，其他的就迎刃而解了。把每个元素看做一个山峰的高度，那么这一排山峰中的每个山谷（即比左右都低的）就可以赋值1，然后它的左右就一定要比它高。所以每个山谷就是一次搜索赋值的起点

比如2,1,3,4,5,那么1就是一个山谷，赋值1，然后再往左右一次赋值得到2,1,2,3,4得到结果是12。

所以在代码中首先找到所有山谷，剩下的比较赋值就行了

public int candy(int[] ratings) {
int len=ratings.length;
if(len ==0 || len ==1){
return len;
}
int A[] = new int[len];//A[i]=0的是山谷
if(ratings[0]>ratings[1]){
A[0]=1;
}
if(ratings[len-1]>ratings[len-2]){
A[len-1]=1;
}
for(int i=1;i<len-1;i++){
if(ratings[i] > ratings[i-1] || ratings[i] > ratings[i+1]){
A[i]=1;
}
}
int V[] = new int[len];

for(int i=0;i<len;i++){
if(A[i] == 0){
V[i]=1;
for(int j=i-1;j>=0&&A[j]==1;j--){//从波谷往左遍历
if(j-1>=0 && ratings[j-1]<ratings[j] && j+1<len && ratings[j+1]<ratings[j]){
V[j]=Math.max(V[j-1], V[j+1])+1;
}
else if(j-1>=0 && ratings[j-1]<ratings[j]){
V[j]=V[j-1]+1;
}
else if(ratings[j+1]<ratings[j]){
V[j]=V[j+1]+1;
}
}
for(int j=i+1;j<len&&A[j]==1;j++){//从波谷往右遍历
if(j-1>=0 && ratings[j-1]<ratings[j] && j+1<len && ratings[j+1]<ratings[j]){
V[j]=Math.max(V[j-1], V[j+1])+1;
}
else if(j-1>=0 && ratings[j-1]<ratings[j]){
V[j]=V[j-1]+1;
}
else if(ratings[j+1]<ratings[j]){
V[j]=V[j+1]+1;
}
}
}
}

int sum = 0;
for(int i=0;i<len;i++){
sum+=V[i];
}
return sum;
}