标签:leetcode
三种解法:推荐第一种和最后一种,
方法一:遍历2次,时间复杂度:O(n),看见复杂度:o(n)
方法三:遍历一次,时间复杂度:o(n), 空间复杂大:o(1)
/* 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. */ #include<iostream> #include<vector> //#include<algorithm> #include <windows.h> using namespace std; #define STOP system("pause"); #if 1 class Solution { public: int candy(vector<int> &ratings) { int len = ratings.size(); int *candy = new int[len]{1};//只有candy[0]=1,others = 0; //vector<int> candy(len, 1); for (int i = 1; i < len; i++){ if (ratings[i] > ratings[i - 1]){ candy[i] = candy[i - 1] + 1; } else candy[i] = 1; } for (int i = len - 2; i >= 0; i--){ if (ratings[i] > ratings[i + 1]){ candy[i] =max(candy[i], candy[i + 1] + 1); } } int res{}; for (int i = 0; i < len; i++){ res += candy[i]; } return res; } }; #elif 0 class Solution { public: int candy(const vector<int>& ratings) { vector<int> f(ratings.size()); int sum = 0; for (int i = 0; i < ratings.size(); ++i) sum += solve(ratings, f, i); return sum; } int solve(const vector<int>& ratings, vector<int>& f, int i) { if (f[i] == 0) { f[i] = 1; if (i > 0 && ratings[i] > ratings[i - 1]) f[i] = max(f[i], solve(ratings, f, i - 1) + 1); if (i < ratings.size() - 1 && ratings[i] > ratings[i + 1]) f[i] = max(f[i], solve(ratings, f, i + 1) + 1); } return f[i]; } }; #elif 0 class Solution { public: int candy(vector<int> &ratings) { // Note: The Solution object is instantiated only once and is reused by each test case. int len = ratings.size(); int nCandyCnt = 1;///Total candies int nSeqLen = 0; /// Continuous ratings descending sequence length int nPreCanCnt = 1; /// Previous child's candy count int nMaxCntInSeq = nPreCanCnt; //if (ratings.begin() != ratings.end()) //for (vector<int>::iterator i = ratings.begin() + 1; i != ratings.end(); i++) for (int i = 1; i < len; i++) { // if r[k]>r[k+1]>r[k+2]...>r[k+n],r[k+n]<=r[k+n+1], // r[i] needs n-(i-k)+(Pre's) candies(k<i<k+n) // But if possible, we can allocate one candy to the child, // and with the sequence extends, add the child's candy by one // until the child's candy reaches that of the prev's. // Then increase the pre's candy as well. // if r[k] < r[k+1], r[k+1] needs one more candy than r[k] // if (ratings[i] < ratings[i - 1]) { //Now we are in a sequence nSeqLen++; if (nMaxCntInSeq == nSeqLen) { //The first child in the sequence has the same candy as the prev //The prev should be included in the sequence. nSeqLen++; } nCandyCnt += nSeqLen; nPreCanCnt = 1; } else { if (ratings[i] > ratings[i - 1]) { nPreCanCnt++; } else { nPreCanCnt = 1; } nCandyCnt += nPreCanCnt; nSeqLen = 0; nMaxCntInSeq = nPreCanCnt; } } return nCandyCnt; } }; #endif void test0(){ vector<int> a{ 0, 1, 3,1,2,3,4,5,6,5,4,3,2,1 }; int r[10]{1, 2}; Solution ss; ss.candy(a); } int main(){ LARGE_INTEGER nFreq; LARGE_INTEGER nBeginTime; LARGE_INTEGER nEndTime; double time; QueryPerformanceFrequency(&nFreq); QueryPerformanceCounter(&nBeginTime); test0(); QueryPerformanceCounter(&nEndTime); time = (double)(nEndTime.QuadPart - nBeginTime.QuadPart) / (double)nFreq.QuadPart; cout << time << endl; STOP; return 0; }
标签:leetcode
原文地址:http://blog.csdn.net/u011409995/article/details/39048673