题目描述
合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。
示例:
输入:
[
1->4->5,
1->3->4,
2->6
]
输出: 1->1->2->3->4->4->5->6
一 思路一 最笨方法
- 将所有节点添加到一个数组中
- 对数组中的节点从小到大进行排序
- 从数组中从小到大依次取出节点，串成链表
图解
image
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- 核心代码如下
- (void)viewDidLoad {
[super viewDidLoad];
// Do any additional setup after loading the view, typically from a nib.
// 构造三个链表
LinkNode *k1 = [[LinkNode alloc] initWithPrev:nil element:@(1) next:nil];
LinkNode *k2 = [[LinkNode alloc] initWithPrev:k1 element:@(4) next:nil];
LinkNode *k3 = [[LinkNode alloc] initWithPrev:k2 element:@(7) next:nil];
LinkNode *k4 = [[LinkNode alloc] initWithPrev:nil element:@(2) next:nil];
LinkNode *k5 = [[LinkNode alloc] initWithPrev:k4 element:@(5) next:nil];
LinkNode *k6 = [[LinkNode alloc] initWithPrev:k5 element:@(8) next:nil];
LinkNode *k7 = [[LinkNode alloc] initWithPrev:nil element:@(3) next:nil];
LinkNode *k8 = [[LinkNode alloc] initWithPrev:k7 element:@(6) next:nil];
LinkNode *k9 = [[LinkNode alloc] initWithPrev:k8 element:@(9) next:nil];
// 方法一
NSArray *linkNodes = @[k1,k4,k7];
LinkNode *k = [self mergeManyLists:linkNodes];
while (k) {
NSLog(@"%@",[k description]);
k = k.next;
}
}
/// 方法一:合并N个有序链表
- (LinkNode *)mergeManyLists:(NSArray *)linkNodes {
if (!linkNodes || linkNodes.count == 0) {
return nil;
}
NSMutableArray *nodes = [NSMutableArray array];
for (LinkNode *node in linkNodes) {
__strong LinkNode *strongNode = node;
while (strongNode != nil) {
[nodes addObject:strongNode];
strongNode = strongNode.next;
}
}
// 排序
NSArray *newNodes = [nodes sortedArrayUsingComparator:^NSComparisonResult(LinkNode *obj1, LinkNode *obj2) {
if (obj1.value > obj2.value) {
return NSOrderedDescending;
}
return NSOrderedAscending;
}];
LinkNode *head = [[LinkNode alloc] init];
LinkNode *cur = head;
for (LinkNode *node in newNodes) {
cur = cur.next = node;
}
return head.next;
}
- 运行结果
2019-11-12 23:12:02.674864+0800 02_MergeTwoLists[30509:1059927] (null)_1_2
2019-11-12 23:12:02.675147+0800 02_MergeTwoLists[30509:1059927] 1_2_3
2019-11-12 23:12:02.675315+0800 02_MergeTwoLists[30509:1059927] 2_3_4
2019-11-12 23:12:02.675583+0800 02_MergeTwoLists[30509:1059927] 3_4_5
2019-11-12 23:12:02.675706+0800 02_MergeTwoLists[30509:1059927] 4_5_6
2019-11-12 23:12:02.675810+0800 02_MergeTwoLists[30509:1059927] 5_6_7
2019-11-12 23:12:02.675941+0800 02_MergeTwoLists[30509:1059927] 6_7_8
2019-11-12 23:12:02.676053+0800 02_MergeTwoLists[30509:1059927] 7_8_9
2019-11-12 23:12:02.676159+0800 02_MergeTwoLists[30509:1059927] 8_9_null
时间复杂度:O(nlogn)
空间复杂度:O(n)
二 思路二 逐一比较
image
- 核心代码如下
/// 方法二 逐一比较
- (LinkNode *)mergeManyLists2:(NSMutableArray<LinkNode *> *)linkNodes {
if (linkNodes == nil || linkNodes.count == 0) {
return nil;
}
LinkNode *head = [[LinkNode alloc] init];
LinkNode *cur = head;
while (true) {
int minIndex = -1;
for (int i = 0; i < linkNodes.count; i++) {
if (linkNodes[i] == nil || linkNodes[i].element == nil) {
continue;
}
if (minIndex == -1 || linkNodes[i].value < linkNodes[minIndex].value) {
minIndex = i;
}
}
if (minIndex == -1) {
break;
}
cur = cur.next = linkNodes[minIndex];
linkNodes[minIndex] = linkNodes[minIndex].next;
}
return head.next;
}
- 测试代码
- (void)viewDidLoad {
[super viewDidLoad];
// Do any additional setup after loading the view, typically from a nib.
LinkNode *nilNode = [[LinkNode alloc] initWithPrev:nil element:nil next:nil];
// 构造三个链表
LinkNode *k1 = [[LinkNode alloc] initWithPrev:nil element:@(1) next:nil];
LinkNode *k2 = [[LinkNode alloc] initWithPrev:k1 element:@(4) next:nil];
LinkNode *k3 = [[LinkNode alloc] initWithPrev:k2 element:@(7) next:nil];
LinkNode *k4 = [[LinkNode alloc] initWithPrev:nil element:@(2) next:nil];
LinkNode *k5 = [[LinkNode alloc] initWithPrev:k4 element:@(5) next:nil];
LinkNode *k6 = [[LinkNode alloc] initWithPrev:k5 element:@(8) next:nil];
LinkNode *k7 = [[LinkNode alloc] initWithPrev:nil element:@(3) next:nil];
LinkNode *k8 = [[LinkNode alloc] initWithPrev:k7 element:@(6) next:nil];
LinkNode *k9 = [[LinkNode alloc] initWithPrev:k8 element:@(9) next:nil];
// 为了需要,每个节点最后是一个空的节点
k3.next = nilNode;
k6.next = nilNode;
k9.next = nilNode;
// 方法二 逐一比较
NSMutableArray *linkNodes = [NSMutableArray arrayWithArray:@[k1,k4,k7]];
LinkNode *k = [self mergeManyLists2:linkNodes];
while (k) {
NSLog(@"%@",[k description]);
k = k.next;
}
}
- 运行结果如下
2019-11-17 10:43:34.914883+0800 02_MergeTwoLists[40077:1519139] (null)_1_2
2019-11-17 10:43:34.915081+0800 02_MergeTwoLists[40077:1519139] 1_2_3
2019-11-17 10:43:34.915177+0800 02_MergeTwoLists[40077:1519139] 2_3_4
2019-11-17 10:43:34.915274+0800 02_MergeTwoLists[40077:1519139] 3_4_5
2019-11-17 10:43:34.915491+0800 02_MergeTwoLists[40077:1519139] 4_5_6
2019-11-17 10:43:34.915623+0800 02_MergeTwoLists[40077:1519139] 5_6_7
2019-11-17 10:43:34.915744+0800 02_MergeTwoLists[40077:1519139] 6_7_8
2019-11-17 10:43:34.915865+0800 02_MergeTwoLists[40077:1519139] 7_8_9
2019-11-17 10:43:34.915981+0800 02_MergeTwoLists[40077:1519139] 8_9_(null)
2019-11-17 10:43:34.916102+0800 02_MergeTwoLists[40077:1519139] 9_(null)_null
三 思路三-逐一两两合并
image
- 核心代码
/// 方法三 逐一两两合并
- (LinkNode *)mergeManyLists3:(NSMutableArray<LinkNode *> *)linkNodes {
if (linkNodes == nil || linkNodes.count == 0) {
return nil;
}
for (int i = 1; i < linkNodes.count; i++) {
linkNodes[0] = [self mergeTwoLists:linkNodes[0] k2:linkNodes[i]];
}
return linkNodes[0];
}
/// 方法一:递归
/// 1.只要是用到递归,首先要搞清楚一个问题,这个递归函数的功能是什么
/// 2.递归基:边界
- (LinkNode *)mergeTwoLists:(LinkNode *)k1 k2:(LinkNode *)k2 {
if (k1 == nil) return k2;
if (k2 == nil) return k1;
if (k1.value <= k2.value) {
k1.next = [self mergeTwoLists:k1.next k2:k2];
return k1;
} else {
k2.next = [self mergeTwoLists:k1 k2:k2.next];
return k2;
}
}
- 测试代码
/// 方法三 逐一两两合并
NSMutableArray *linkNodes = [NSMutableArray arrayWithArray:@[k1,k4,k7]];
LinkNode *k = [self mergeManyLists3:linkNodes];
while (k) {
NSLog(@"%@",[k description]);
k = k.next;
}
- 运行结果
2019-11-17 11:19:14.687574+0800 02_MergeTwoLists[40935:1548483] null_1_2
2019-11-17 11:19:14.687798+0800 02_MergeTwoLists[40935:1548483] 1_2_3
2019-11-17 11:19:14.687948+0800 02_MergeTwoLists[40935:1548483] 2_3_4
2019-11-17 11:19:14.688098+0800 02_MergeTwoLists[40935:1548483] 3_4_5
2019-11-17 11:19:14.688245+0800 02_MergeTwoLists[40935:1548483] 4_5_6
2019-11-17 11:19:14.688401+0800 02_MergeTwoLists[40935:1548483] 5_6_7
2019-11-17 11:19:14.688515+0800 02_MergeTwoLists[40935:1548483] 6_7_8
2019-11-17 11:19:14.688620+0800 02_MergeTwoLists[40935:1548483] 7_8_9
2019-11-17 11:19:14.688727+0800 02_MergeTwoLists[40935:1548483] 8_9_null
四 思路四-优先级队列（小顶堆）
image
五 思路五-分治策略
image
- 核心代码
/// 方法五 - 分治策略
- (LinkNode *)mergeManyLists5:(NSMutableArray<LinkNode *> *)linkNodes {
if (linkNodes == nil || linkNodes.count == 0) {
return nil;
}
int step = 1;
while (step < linkNodes.count) {
int nextStep = step << 1;
for (int i = 0; i + step < linkNodes.count; i += nextStep) {
linkNodes[i] = [self mergeTwoLists:linkNodes[i] k2:linkNodes[i + step]];
}
step = nextStep;
}
return linkNodes[0];
}
- 运行结果
2019-11-17 21:57:36.968482+0800 02_MergeTwoLists[54576:1953555] null_1_2
2019-11-17 21:57:36.968663+0800 02_MergeTwoLists[54576:1953555] 1_2_3
2019-11-17 21:57:36.968778+0800 02_MergeTwoLists[54576:1953555] 2_3_4
2019-11-17 21:57:36.968890+0800 02_MergeTwoLists[54576:1953555] 3_4_5
2019-11-17 21:57:36.969008+0800 02_MergeTwoLists[54576:1953555] 4_5_6
2019-11-17 21:57:36.969148+0800 02_MergeTwoLists[54576:1953555] 5_6_7
2019-11-17 21:57:36.969267+0800 02_MergeTwoLists[54576:1953555] 6_7_8
2019-11-17 21:57:36.969376+0800 02_MergeTwoLists[54576:1953555] 7_8_9
2019-11-17 21:57:36.969485+0800 02_MergeTwoLists[54576:1953555] 8_9_null
作为一个开发者，有一个学习的氛围跟一个交流圈子特别重要，这是一个我的iOS交流群：413038000，不管你是大牛还是小白都欢迎入驻 ，分享BAT,阿里面试题、面试经验，讨论技术， 大家一起交流学习成长！
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iOS开发——最新 BAT面试题合集（持续更新中）
作者：路飞_Luck
链接：https://www.jianshu.com/p/43ecb9d85461
来源：简书
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