题目地址：here
题目大意：几个小孩站一排，每个小孩有个等级值，现在给小孩发糖，发的时候要遵守2个规则：（1）每个小孩至少一颗糖（2）两个相邻的小孩中，等级大的小孩一定比等级小的小孩糖多，求发糖的数目的最小值。
本文提供两个算法，第一个是我自己做题时用的，第二个是网上看题解时看到的
算法1：该算法采用分治发，把小孩平均分左右两拨，求得两拨小孩的最小值分配方案后，还得看分界线出是否满足规则。我们用L[end]表示左边一拨小孩的右边界，R[start]表示右边一拨小孩的左边界，处理分界线出的情况分以下两种：
（1）如果L[end]等级比R[start]高，但是糖数少或相等，那么就从左边一拨小孩的右边界依次调整糖数以满足规则；
（2）如果L[end]等级比R[start]低，但是糖数多或相等，那么就从右边一拨小孩的左边界依次调整糖数以满足规则；
该算法时间复杂度为O(N*lgN)
算法1代码如下：
1 class Solution { 2 public: 3 int candy(vector<int> &ratings) 4 { 5 // Note: The Solution object is instantiated only once and is reused by each test case. 6 int *candyNum = new int[ratings.size()];//每个小孩的糖数目 7 memset(candyNum,0,sizeof(int)*ratings.size()); 8 int result = candyRecursive(ratings, 0, ratings.size()-1, candyNum); 9 delete []candyNum; 10 return result; 11 } 12 13 int candyRecursive(vector<int> &ratings, int startloc, int endloc, 14 int candyNum[]) 15 { 16 if(startloc == endloc) 17 { 18 candyNum[startloc] = 1; 19 return 1; 20 } 21 int middle = (startloc + endloc)/2; 22 int rightCandy = candyRecursive(ratings, middle+1, endloc, candyNum); 23 int leftCandy = candyRecursive(ratings, startloc, middle, candyNum); 24 if(ratings[middle+1] > ratings[middle]) 25 { 26 int tmp = candyNum[middle+1] - candyNum[middle]; 27 if(tmp <= 0) 28 { 29 tmp *= -1; 30 candyNum[middle+1] += (tmp+1); 31 rightCandy += (tmp+1); 32 for(int i = middle+2; i <= endloc; i++) 33 { 34 if(ratings[i] > ratings[i-1] && candyNum[i] <= candyNum[i-1]) 35 { 36 int foo = candyNum[i-1] - candyNum[i] + 1; 37 candyNum[i] += foo; 38 rightCandy += foo; 39 } 40 else break; 41 } 42 } 43 } 44 else if(ratings[middle+1] < ratings[middle]) 45 { 46 int tmp = candyNum[middle+1] - candyNum[middle]; 47 if(tmp >= 0) 48 { 49 candyNum[middle] += (tmp+1); 50 leftCandy += (tmp+1); 51 for(int i = middle-1; i >= startloc; i--) 52 { 53 if(ratings[i] > ratings[i+1] && candyNum[i] <= candyNum[i+1]) 54 { 55 int foo = (candyNum[i+1] - candyNum[i] + 1); 56 candyNum[i] += foo; 57 leftCandy += foo; 58 } 59 else break; 60 } 61 } 62 } 63 return leftCandy + rightCandy; 64 } 65 };
算法2：初始化所有小孩糖数目为1，从前往后扫描，如果第i个小孩等级比第i-1个高，那么i的糖数目等于i-1的糖数目+1；从后往前扫描，如果第i个的小孩的等级比i+1个小孩高,但是糖的数目却小或者相等，那么i的糖数目等于i+1的糖数目+1。
该算法时间复杂度为O(N)
算法2代码如下：
1 class Solution { 2 public: 3 int candy(vector<int> &ratings) 4 { 5 // Note: The Solution object is instantiated only once and is reused by each test case. 6 int *candyNum = new int[ratings.size()];//每个小孩的糖数目 7 for(int i = 0; i < ratings.size(); i++) 8 candyNum[i] = 1; 9 for(int i = 1; i < ratings.size(); i++) 10 if(ratings[i] > ratings[i-1]) 11 candyNum[i] = candyNum[i-1] + 1; 12 for(int i = ratings.size()-2; i>=0; i--) 13 if(ratings[i] > ratings[i+1] && candyNum[i] <= candyNum[i+1]) 14 candyNum[i] = candyNum[i+1] + 1; 15 int result = 0; 16 for(int i = 0; i < ratings.size(); i++) 17 result += candyNum[i]; 18 delete []candyNum; 19 return result; 20 } 21 };
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