forked from zhedahht/CodingInterviewChinese2
-
Notifications
You must be signed in to change notification settings - Fork 0
/
LastNumberInCircle.cpp
131 lines (103 loc) · 2.61 KB
/
LastNumberInCircle.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者:何海涛
//==================================================================
// 面试题62:圆圈中最后剩下的数字
// 题目:0, 1, …, n-1这n个数字排成一个圆圈,从数字0开始每次从这个圆圈里
// 删除第m个数字。求出这个圆圈里剩下的最后一个数字。
#include <cstdio>
#include <list>
using namespace std;
// ====================方法1====================
int LastRemaining_Solution1(unsigned int n, unsigned int m)
{
if(n < 1 || m < 1)
return -1;
unsigned int i = 0;
list<int> numbers;
for(i = 0; i < n; ++ i)
numbers.push_back(i);
list<int>::iterator current = numbers.begin();
while(numbers.size() > 1)
{
for(int i = 1; i < m; ++ i)
{
current ++;
if(current == numbers.end())
current = numbers.begin();
}
list<int>::iterator next = ++ current;
if(next == numbers.end())
next = numbers.begin();
-- current;
numbers.erase(current);
current = next;
}
return *(current);
}
// ====================方法2====================
int LastRemaining_Solution2(unsigned int n, unsigned int m)
{
if(n < 1 || m < 1)
return -1;
int last = 0;
for (int i = 2; i <= n; i ++)
last = (last + m) % i;
return last;
}
// ====================测试代码====================
void Test(const char* testName, unsigned int n, unsigned int m, int expected)
{
if(testName != nullptr)
printf("%s begins: \n", testName);
if(LastRemaining_Solution1(n, m) == expected)
printf("Solution1 passed.\n");
else
printf("Solution1 failed.\n");
if(LastRemaining_Solution2(n, m) == expected)
printf("Solution2 passed.\n");
else
printf("Solution2 failed.\n");
printf("\n");
}
void Test1()
{
Test("Test1", 5, 3, 3);
}
void Test2()
{
Test("Test2", 5, 2, 2);
}
void Test3()
{
Test("Test3", 6, 7, 4);
}
void Test4()
{
Test("Test4", 6, 6, 3);
}
void Test5()
{
Test("Test5", 0, 0, -1);
}
void Test6()
{
Test("Test6", 4000, 997, 1027);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
return 0;
}