1、我去，这个竟然是前两天百度的面试题，不是最优就挂掉，看来面试官也在网上搜新题啊。
#include <iostream> #include <queue> #include <algorithm> #include <vector> using namespace std; class ChocolateBar { public: int maxLength(string letters) { int res = 0; vector<int> last(256,-1); int l_size = letters.size(); int cur_min=0; for (int i = 0; i < l_size; i++) { int cur=letters[i]; cur_min=min(cur_min+1,i-last[cur]); last[cur]=i; res=max(cur_min,res); } return res; } };
2、感觉像是在考我们对队列的应用。
1 #include <iostream> 2 #include <queue> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 7 class XorTravelingSalesman { 8 public: 9 int maxProfit(vector<int> cityValues, vector<string> roads) { 10 int val_size = cityValues.size(); 11 12 vector<int> in(val_size, 0); 13 vector<vector<int> > judge(val_size, vector<int>(1024, 0)); 14 15 queue<int> my_queue; 16 my_queue.push(0); 17 in[0] = 1; 18 judge[0][cityValues[0]] = 1; 19 20 int res = 0; 21 22 while (!my_queue.empty()) { 23 int cur = my_queue.front(); 24 my_queue.pop(); 25 in[cur] = 0; 26 int i = 0; 27 int cur_max = 0; 28 for (i = 0; i < 1024; i++) { 29 if (judge[cur][i]) { 30 cur_max = i; 31 res = max(res, cur_max); 32 int k; 33 for (k = 0; k < roads[0].size(); k++) { 34 if (roads[cur][k] == 'Y' 35 && (0 == judge[k][cur_max ^ cityValues[k]])) { 36 if (0 == in[k]) { 37 in[k] = 1; 38 my_queue.push(k); 39 } 40 judge[k][cityValues[k] ^ cur_max] = 1; 41 } 42 } 43 } 44 } 45 res = max(res, cur_max); 46 47 } 48 return res; 49 } 50 };
3、只要分清两种可能就可以了，最小的数字只能由之前的最小的非0开头的数字和本数字结合，或者之前最小的0开头的数字和本数字结合。
1 #include <iostream> 2 #include <queue> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 7 class LeftRightDigitsGame { 8 public: 9 string minNumber(string digits) { 10 vector<string> cur; 11 vector<string>::iterator it_cur; 12 int i, j; 13 14 int size_digits = digits.length(); 15 string zero, non_zero; 16 string tmp; 17 tmp += digits[0]; 18 cur.push_back(tmp); 19 for (i = 1; i < size_digits; i++) { 20 sort(cur.begin(), cur.end()); 21 int zero_exi = 0; 22 int non_zero_exi = 0; 23 for (j = cur.size() - 1; j >= 0; j--) {//倒序保证是最小的 24 if (cur[j][0] == '0') { 25 zero = cur[j]; 26 zero_exi = 1; 27 } else { 28 non_zero = cur[j]; 29 non_zero_exi = 1; 30 } 31 } 32 cur.clear();//为下一次循环做准备 33 if (zero_exi) { 34 cur.push_back(zero + digits[i]); 35 cur.push_back(digits[i] + zero); 36 } 37 if (non_zero_exi) { 38 cur.push_back(non_zero + digits[i]); 39 cur.push_back(digits[i] + zero); 40 } 41 } 42 sort(cur.begin(), cur.end()); 43 string res; 44 for (j = cur.size() - 1; j >= 0; j--) { 45 if (cur[j][0] != '0') {//挑出最小的 46 res = cur[j]; 47 } 48 } 49 return res; 50 } 51 };