标签：style blog io for ar 问题 div line
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:
What is the minimum candies you must give?
分析: 主要问题是可以从左升序或者从右升序，如何取大值。
方法一：从左从右分别计算一次，对值校正。并计算出最大值。要保存中间初步的值，空间复杂度:O(n)
class Solution {
public:
int candy(vector<int> &ratings) {
int n = ratings.size();
if(n == 0) return 0;
int totalCandy = 0;
int *getCandy = new int[n];
getCandy[0] = 1;
for(int i = 1; i < n; ++i)
if(ratings[i] > ratings[i-1]) getCandy[i] = getCandy[i-1] + 1;
else getCandy[i] = 1;
totalCandy += getCandy[n-1];
for(int i = n-1; i > 0; --i) {
if(ratings[i-1] > ratings[i]) getCandy[i-1] = max(getCandy[i-1], getCandy[i]+1);
totalCandy += getCandy[i-1];
}
delete[] getCandy;
return totalCandy;
}
};
方法二：从一个方向（此题从左），设置两个变量，通过计算升序长度，降序长度确定精确值。空间复杂度:O(1)
class Solution {
public:
int candy(vector<int> &ratings) {
int LA = 0, LD = 0; // 利用降序长度和升序长度，来求结果。
bool decend = false;
int totalCandy = ratings.size();
for(int i = 1; i < ratings.size(); ++i) {
if(ratings[i-1] < ratings[i]) {
if(LA == 0 || decend == true) LA = 2;//考虑情况：之前为降序，重新设置升序长度
else ++LA;
LD = 0; decend = false; //升序时不需要知道之前降序长度。
totalCandy += (LA - 1);
}else if(ratings[i-1] > ratings[i]) {
decend = true;
if(LD == 0) LD = 2;
else ++LD;
if(LD <= LA) totalCandy += (LD - 2);
else totalCandy += (LD - 1);
}else {
LA = LD = 0; // 出现相同字符，则从第二个重复字符重新开始
}
}
return totalCandy;
}
};
标签：style blog io for ar 问题 div line
原文地址：http://www.cnblogs.com/liyangguang1988/p/3902305.html
