1. Sort
郭至軒（KuoE0）
KuoE0.tw@gmail.com
KuoE0.ch

2. Attribution-ShareAlike 3.0 Unported
(CC BY-SA 3.0)
http://creativecommons.org/licenses/by-sa/3.0/
Latest update: Feb 27, 2013

3. Bubble Sort
氣泡排序法

4. Algorithm
1. 從第一個元素開始比較每一對相鄰元素
2. 若第一個元素大於第二個元素則交換
3. 每次遞減需要比較的元素對直到不需比
較

5. Origin
10 2 7 9 1
10 2 7 9 1

6. 1 st Iteration
10 2 7 9 1

7. 1 st Iteration
10 2 7 9 1

8. 1 st Iteration
2
10 2
10 7 9 1

9. 1 st Iteration
2
10 2
10 7 9 1

10. 1 st Iteration
2
10 2
7 7
10 9 1

11. 1 st Iteration
2
10 2
7 7
10 9 1

12. 1 st Iteration
2
10 2
7 7
9 9
10 1

13. 1 st Iteration
2
10 2
7 7
9 9
10 1

14. 1 st Iteration
2
10 2
7 7
9 1
9 1
10

15. 1 st Iteration
2 7 9 1 10
2
10 2
7 7
9 1
9 1
10

16. 2 nd Iteration
2 7 9 1 10

17. 2 nd Iteration
2 7 9 1 10

18. 2 nd Iteration
2 7 9 1 10

19. 2 nd Iteration
2 7 9 1 10

20. 2 nd Iteration
2 7 9 1 10

21. 2 nd Iteration
2 7 9 1 10

22. 2 nd Iteration
2 7 1
9 1
9 10

23. 2 nd Iteration
2 7 1 9 10
2 7 1
9 1
9 10

24. 3 rd Iteration
2 7 1 9 10

25. 3 rd Iteration
2 7 1 9 10

26. 3 rd Iteration
2 7 1 9 10

27. 3 rd Iteration
2 7 1 9 10

28. 3 rd Iteration
2 1
7 1
7 9 10

29. 3 rd Iteration
2 1 7 9 10
2 1
7 1
7 9 10

30. 4 th Iteration
2 1 7 9 10

31. 4 th Iteration
2 1 7 9 10

32. 4 th Iteration
1
2 1
2 7 9 10

33. 4 th Iteration
1 2 7 9 10
1
2 1
2 7 9 10

34. Result
1 2 7 9 10
1 2 7 9 10

35. Source Code
for ( int i = 0; i < n; ++i )
for ( int j = 0; j < n - 1 - i; ++j )
if ( A[ j ] > A[ j + 1 ] )
swap( A[ j ], A[ j + 1 ] );

36. Time Complexity
for ( int i = 0; i < n; ++i )
for ( int j = 0; j < n - 1 - i; ++j )
if ( A[ j ] > A[ j + 1 ] )
swap( A[ j ], A[ j + 1 ] );
最大循環次數略估
第 1 層迴圈 n
第 2 層迴圈 n

37. n×n= n 2
Time Complexity: O(n2)

38. 從範例中可發現...
耗費許多時間檢查有序數對

39. Merge Sort
合併排序法

40. Algorithm
採用 divide & conquer 策略

41. Divide
將當前數列對半切割遞迴進行
直到 剩一個元素

42. 2 9 4 3 8 7 5 1 6
2 9 4 3 8 7 5 1 6
2 9 4 3 8 7 5 1 6
2 9 4 3 8 7 5 1 6
2 9

43. Conquer
1. 利用兩個指標指向兩個有序數列 A 與 B
2. 比較指標指向的數值
3. 將較小的數值放入新的數列 C，並將該指標
指向下一數值
4. 直到某一指標指向數列結尾，將另一數列剩
餘的數值放都入數列 C

44. Merge Two Sequence
2 3 4 8 9 1 5 6 7

45. Merge Two Sequence
2 3 4 8 9 5 6 7
1

46. Merge Two Sequence
3 4 8 9 5 6 7
1 2

47. Merge Two Sequence
4 8 9 5 6 7
1 2 3

48. Merge Two Sequence
8 9 5 6 7
1 2 3 4

49. Merge Two Sequence
8 9 6 7
1 2 3 4 5

50. Merge Two Sequence
8 9 7
1 2 3 4 5 6

51. Merge Two Sequence
8 9
1 2 3 4 5 6 7

52. Merge Two Sequence
9
1 2 3 4 5 6 7 8

53. Merge Two Sequence
1 2 3 4 5 6 7 8 9

54. 1 2 3 4 5 6 7 8 9
2 3 4 8 9 1 5 6 7
2 4 9 3 8 5 7 1 6
2 9 4 3 8 7 5 1 6
2 9

55. Source Code
int mergeSort( int L, int R ) {
if ( R - L == 1 )
return;
int mid = ( R + L ) / 2, C[ R - L ];
mergeSort( L, mid );
mergeSort( mid, R );
for ( int p1 = L, p2 = mid, p = 0; p < R - L; ++p ) {
if ( ( p1 != mid && A[ p1 ] < A[ p2 ] ) || p2 == R )
C[ p ] = A[ p1++ ];
else
C[ p ] = A[ p2++ ];
}
for ( int i = L; i < R; ++i )
A[ i ] = C[ i - L ];
}

56. Time Complexity
total time

57. Time Complexity
total time
n
2 * (n/2) = n
...
n * (n/n) = n

58. Best height
}
log2n

59. Time Complexity: O(nlog2n)

60. Quick Sort
快速排序法

61. Algorithm
採用 divide & conquer 策略

62. Divide
從數列中挑出一個元素作為 pivot，利
用 pivot 將原數列分為兩個子數列：
• 所有元素小於 pivot 的數列 A
• 所有元素大於 pivot 的元素的數列 B
• 等於 pivot 的元素可放置在任一個中

63. 2 9 4 3 8 7 5 1 6

64. 2 9 4 3 8 7 5 1 6

65. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8

66. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8

67. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6

68. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6

69. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6
2

70. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6
2

71. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6
2 5 6

72. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6
2 5 6

73. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6
2 5 6
5

74. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6
2 5 6
5

75. 2 9 4 3 8 7 5 1 6
2 4 3 5 1 6 9 8
2 1 4 5 6 9
2 5 6
5

76. Conquer
由於子數列已被 pivot 分為兩段，一段
小於等於 pivot 的數列 A，另一段大於
等於 pivot 的數列 B。

77. Conquer
由於子數列已被 pivot 分為兩段，一段
小於等於 pivot 的數列 A，另一段大於
等於 pivot 的數列 B。
A C B

78. 2

79. 1
2

80. 1 2
2

81. 1 2
2 6
5

82. 1 2
2 5 6
5

83. 1 2 4
2 5 6
5

84. 1 2 4 5 6
2 5 6
5

85. 3
1 2 4 5 6
2 5 6
5

86. 1 2 3 4 5 6
1 2 4 5 6
2 5 6
5

87. 1 2 3 4 5 6 8
1 2 4 5 6 9
2 5 6
5

88. 1 2 3 4 5 6 8 9
1 2 4 5 6 9
2 5 6
5

89. 7
1 2 3 4 5 6 8 9
1 2 4 5 6 9
2 5 6
5

90. 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 8 9
1 2 4 5 6 9
2 5 6
5

91. 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 8 9
1 2 4 5 6 9
2 5 6
5

92. Source Code
void quickSort( int L, int R ) {
if ( R - L <= 1 )
return;
int pivot = N[ L ], p1 = L + 1, p2 = R - 1;
do {
while ( N[ p1++ ] <= pivot )
;
while ( N[ p2-- ] > pivot )
;
if ( p1 < p2 )
swap( N[ p1 ], N[ p2 ] );
} while ( p1 < p2 );
quickSort( L + 1, p1 );
quickSort( p1, R );
for ( int i = L + 1; i < p1; ++i )
swap( N[ i - 1 ], N[ i ] );
}

93. How to Divide
4 1 9 7 5 8 2 3 6

94. How to Divide
4 1 9 7 5 8 2 3 6

95. How to Divide
4 1 9 7 5 8 2 3 6

96. How to Divide
4 1 9 7 5 8 2 3 6

97. How to Divide
4 1 3 7 5 8 2 9 6

98. How to Divide
4 1 3 7 5 8 2 9 6

99. How to Divide
4 1 3 7 5 8 2 9 6

100. How to Divide
4 1 3 2 5 8 7 9 6

101. How to Divide
4 1 3 2 5 8 7 9 6

102. How to Divide
4 1 3 2 5 8 7 9 6

103. How to Divide
4 1 3 2 5 8 7 9 6
4 1 3 2 5 8 7 9 6

104. How to Conquer
4 1 2 3 5 6 7 8 9

105. How to Conquer
1 4 2 3 5 6 7 8 9

106. How to Conquer
1 2 4 3 5 6 7 8 9

107. How to Conquer
1 2 3 4 5 6 7 8 9

108. In-place Version
void quickSort( int L, int R ) {
if ( R - L <= 1 )
return;
int pivot = N[ R - 1 ], p = L;
for ( int i = L; i < R - 1; ++i ) {
if ( N[ i ] <= pivot ) {
swap( N[ i ], N[ p ] );
++p;
}
}
swap( N[ R - 1 ], N[ p ] );
quickSort( L, p );
quickSort( p + 1, R );
}

109. How to Divide
2 9 4 3 8 7 5 1 6

110. How to Divide
2 9 4 3 8 7 5 1 6

111. How to Divide
2 9 4 3 8 7 5 1 6

112. How to Divide
2 9 4 3 8 7 5 1 6

113. How to Divide
2 9 4 3 8 7 5 1 6

114. How to Divide
2 9 4 3 8 7 5 1 6

115. How to Divide
2 9 4 3 8 7 5 1 6

116. How to Divide
2 4
9 9
4 3 8 7 5 1 6

117. How to Divide
2 4
9 9
4 3 8 7 5 1 6

118. How to Divide
2 4
9 9
4 3 8 7 5 1 6

119. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

120. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

121. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

122. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

123. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

124. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

125. How to Divide
2 4
9 3
4 9
3 8 7 5 1 6

126. How to Divide
2 4
9 3
4 5
3 8 7 9
5 1 6

127. How to Divide
2 4
9 3
4 5
3 8 7 9
5 1 6

128. How to Divide
2 4
9 3
4 5
3 8 7 9
5 1 6

129. How to Divide
2 4
9 3
4 5
3 8
1 7 9
5 8
1 6

130. How to Divide
2 4
9 3
4 5
3 1
8 7 9
5 8
1 6

131. How to Divide
2 4
9 3
4 5
3 1
8 6
7 9
5 8
1 7
6

132. How to Divide
2 4
9 3
4 5
3 1
8 6
7 9
5 8
1 7
6
2 4 3 5 1 9 8 7

133. How to Conquer
6
1 2 3 4 5 7 8 9

134. How to Conquer
1 2 3 4 5 6 7 8 9
1 2 3 4 5 7 8 9

135. Time Complexity
total time

136. Time Complexity
total time
n
2 * (n/2) = n
...
n * (n/n) = n

137. Best height
}
log2n

138. Worst height
}
n

139. Time Complexity:
O(nlog2n) ~ O(n 2)

140. Average Time Complexity:
O(nlog2n)

141. Builtin Function
如果你純粹想要排序罷了...

142. qsort (C/C++)
• include <stdlib.h> or <cstdlib>
• compare function
void qsort (void* base, size_t num, size_t
size, int (*compar)(const void*,const void*));

143. void qsort (void* base, size_t num, size_t
size, int (*compar)(const void*,const void*));
base: 指向欲排序列表起始位置之指標
num: 欲排序的元素數量
size: 各元素大小
compar: 比較函式的函式指標

144. Compare Function
Function prototype:
int function_name(const void* p1, const void* p2);
return value means
less than 0 p1 < p2
equal to 0 p1 == p2
greater than 0 p1 > p2

145. Example
int cmp( const void* p1, const void* p2 ) {
return *(int*)p1 - *(int*)p2;
}
int main() {
int n, N[ 10010 ];
while ( scanf( “%d”, &n ) != EOF ) {
int x;
for ( int i = 0; i < n; ++i ) {
scanf( “%d”, &x );
N[ i ] = x;
}
qsort( N, n, sizeof( int ), cmp );
}
return 0;
}

146. sort (STL)
• include <algorithm>
• compare function for customized
behavior
template <class RandomAccessIterator>
void sort (RandomAccessIterator first,
RandomAccessIterator last);
template <class RandomAccessIterator, class Compare>
void sort (RandomAccessIterator first,
RandomAccessIterator last, Compare comp);

147. Sort by operator <
template <class RandomAccessIterator>
void sort (RandomAccessIterator first,
RandomAccessIterator last);
first: 指向欲排序列表起始位置之 iterator
last: 指向欲排序列表結尾位置之 iterator
指標可被轉型為 iterator!
依照該資料型別的小於運算子 (<) 作為排序依據！

148. Customized Data Type
Function prototype for operator <:
bool operator< ( const type_name &p ) const;
return value means
TRUE this < p
FALSE this >= p

149. Example (builtin type)
int main() {
int n, N[ 10010 ];
while ( scanf( “%d”, &n ) != EOF ) {
int x;
for ( int i = 0; i < n; ++i ) {
scanf( “%d”, &x );
N[ i ] = x;
}
sort( N, N + n );
}
return 0;
}

150. Example (custom type)
struct T {
int x, y;
bool operator< ( const T &p ) const {
return x == p.x ? x < p.x : y < p.y;
}
};
T pt[ 10010 ];
int main() {
int n;
while ( scanf( “%d”, &n ) != EOF ) {
int x, y;
for ( int i = 0; i < n; ++i ) {
scanf( “%d %d”, &x, &y );
pt[ i ].x = x, pt[ i ].y = y;
}
sort( pt, pt + n );
}
return 0;
}

151. Sort by Compare Function
template <class RandomAccessIterator, class Compare>
void sort (RandomAccessIterator first,
RandomAccessIterator last, Compare comp);
first: 指向欲排序列表起始位置之 iterator
last: 指向欲排序列表結尾位置之 iterator
comp: 比較函式
指標可被轉型為 iterator!

152. Customized Data Type
Function prototyp:
bool function_name ( type_name p1, type_name p2 );
return value means
TRUE p1 < p2
FALSE p1 >= p2

153. Example (descending)
bool descending( int p1, int p2 ) {
return p1 >= p2;
}
int main() {
int n, N[ 10010 ];
while ( scanf( “%d”, &n ) != EOF ) {
int x;
for ( int i = 0; i < n; ++i ) {
scanf( “%d”, &x );
N[ i ] = x;
}
sort( N, N + n, descending );
}
return 0;
}

154. Reference
• 冒泡排序 - 維基百科，自由的百科全書
• 歸併排序 - 維基百科，自由的百科全書
• 快速排序 - 維基百科，自由的百科全書
• qsort - C++ Reference
• sort - C++ Reference

155. Practice Now
10810 - Ultra-QuickSort

156. Thank You for Your
Listening.