快排
- 先從數(shù)列中取出一個(gè)數(shù)作為基準(zhǔn)數(shù)
- 區(qū)分過(guò)程，將比這個(gè)數(shù)大的數(shù)全部放到它的右邊，小于或等于它的數(shù)全放到左邊
- 再對(duì)左右區(qū)間重復(fù)第二步，直到各區(qū)間只有一個(gè)數(shù)
- (void)kuaipai:(NSMutableArray *)arr left:(NSInteger)left right:(NSInteger)right{
if (left >= right) {
return;
}
NSInteger i = left;
NSInteger j = right;
NSInteger key = [arr[left] integerValue];
while (i < j) {
while (i < j && key <= [arr[j] integerValue]) {
j --;
}
arr[i] = arr[j];
while (i < j && key >= [arr[i] integerValue]) {
i ++;
}
arr[j] = arr[i];
}
arr[i] = [NSNumber numberWithInteger:key];
[self kuaipai:arr left:left right:i-1];
[self kuaipai:arr left:i+1 right:right];
}
冒泡排序
int m[]=[1,2,3,7,4,9];
int i,j,temp;
for (i=m.count; i>=0; i++) {
for (j=0; j<=i; j++) {
if (m[j+1] < m[j]) {
temp = m[j];
m[j] = m[j=1];
m[j+1] = temp;
]
}
}
二分查找
不用中間變量交換 a, b
a = 1 b = 2
a = a+b; a=3 b=2
b = a-b; a=3 b=1
a = a-b; a=2 b=1
a = 1 b = 2
a = a^b;
b = a^b;
a = a^b;
a = 1 b = 2
a = a*b; a=2 b=2
b = a/b; a=2 b=1
a = a/b; a=2 b=1
單鏈表反轉(zhuǎn)
循環(huán)算法
class Node {
int value;
Node next;
}
Node reverseList(Node head) {
Node p1, p2, p3;
if (head == null || hade.next == null) {
return head;
}
p1 = head;
p2 = head.next;
while (p2!=null){
p3 = p2.next;
p2.next = p1;
p1 = p2;
p2 = p3;
}
head.next = null;
head = p1;
return head;
}
遞歸算法
class Node {
int value;
Node next;
}
public Node reverse(Node head) {
if (head == null || head.next == null) {
return head;
}
Node nextNode = head.next;
head.next = null;
Node reverseRest = reverse(nextNode);
nextNode.next = head;
return reverseRest;
}
