Longest Increasing Subsequence
label: dp
Given an unsorted array of integers, find the length of longest increasing subsequence.
Example:
Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Note:
- There may be more than one LIS combination, it is only necessary for you to return the length.
- Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?
方法一:dp
dp[n] = max{1,dp[i]+1}
代码:
def lengthOfLIS(nums):
"""
:type nums: List[int]
:rtype: int
"""
length = len(nums)
dp = [0]*length
for i in range(length):
maxnum = 1
for j in range(0,i):
if nums[i]>nums[j]:
maxnum = max(maxnum,dp[j]+1)
dp[i] = maxnum
r = 0
for i in range(length):
r = max(r,dp[i])
print(r)
return r
方法二:来自leetcode
tails
is an array storing(储存) the smallest tail of all increasing subsequences with length i+1
in tails[i]
.
For example, say we have nums = [4,5,6,3]
, then all the available increasing subsequences are:
定义一个 tails 数组,其中 tails[i] 存储长度为 i + 1 的最长递增子序列的最后一个元素。如果有多个长度相等的最长递增子序列,那么 tails[i] 就取最小值。例如对于数组 [4,5,6,3],有
len = 1 : [4], [5], [6], [3] => tails[0] = 3
len = 2 : [4, 5], [5, 6] => tails[1] = 5
len = 3 : [4, 5, 6] => tails[2] = 6
We can easily prove(证明) that tails is a increasing array. Therefore it is possible to do a binary search in tails array to find the one needs update.
Each time we only do one of the two:
(1) if x is larger than all tails, append it, increase the size by 1
(2) if tails[i-1] < x <= tails[i], update tails[i]
Doing so will maintain the tails invariant. The the final answer is just the size.
def lengthOfLIS(self, nums):
tails = [0] * len(nums)
size = 0
for x in nums:
i, j = 0, size
while i != j:
m = (i + j) / 2
if tails[m] < x:
i = m + 1
else:
j = m
tails[i] = x
size = max(i + 1, size)
return size
const lengthOfLIS = function(nums) {
let tails = [],
len = 0,
first,
last,
mid;
for (let i = 0, ii = nums.length; i < ii; ++i) {
first = 0;
last = len;
while (first < last) {
mid = parseInt((first + last) / 2);
console.log(nums[i], tails[mid],nums[i] > tails[mid]);
if (nums[i] > tails[mid]) {
first = mid + 1;
} else {
last = mid;
}
}
tails[last] = nums[i];
if (last === len) {
++len;
}
}
return len;
};
一个通俗易懂的解释,来自Felix021
假设存在一个序列d[1..9] = 2 1 5 3 6 4 8 9 7,可以看出来它的LIS长度为5。 下面一步一步试着找出它。 我们定义一个序列B,然后令 i = 1 to 9 逐个考察这个序列。 此外,我们用一个变量Len来记录现在最长算到多少了
首先,把d[1]有序地放到B里,令B[1] = 2,就是说当只有1一个数字2的时候,长度为1的LIS的最小末尾是2。这时Len=1
然后,把d[2]有序地放到B里,令B[1] = 1,就是说长度为1的LIS的最小末尾是1,d[1]=2已经没用了,很容易理解吧。这时Len=1
接着,d[3] = 5,d[3]>B[1],所以令B[1+1]=B[2]=d[3]=5,就是说长度为2的LIS的最小末尾是5,很容易理解吧。这时候B[1..2] = 1, 5,Len=2
再来,d[4] = 3,它正好加在1,5之间,放在1的位置显然不合适,因为1小于3,长度为1的LIS最小末尾应该是1,这样很容易推知,长度为2的LIS最小末尾是3,于是可以把5淘汰掉,这时候B[1..2] = 1, 3,Len = 2
继续,d[5] = 6,它在3后面,因为B[2] = 3, 而6在3后面,于是很容易可以推知B[3] = 6, 这时B[1..3] = 1, 3, 6,还是很容易理解吧? Len = 3 了噢。
第6个, d[6] = 4,你看它在3和6之间,于是我们就可以把6替换掉,得到B[3] = 4。B[1..3] = 1, 3, 4, Len继续等于3
第7个, d[7] = 8,它很大,比4大,嗯。于是B[4] = 8。Len变成4了
第8个, d[8] = 9,得到B[5] = 9,嗯。Len继续增大,到5了。
最后一个, d[9] = 7,它在B[3] = 4和B[4] = 8之间,所以我们知道,最新的B[4] =7,B[1..5] = 1, 3, 4, 7, 9,Len = 5。 于是我们知道了LIS的长度为5。
!!!!! 注意。这个1,3,4,7,9不是LIS,它只是存储的对应长度LIS的最小末尾。有了这个末尾,我们就可以一个一个地插入数据。虽然最后一个d[9] = 7更新进去对于这组数据没有什么意义,但是如果后面再出现两个数字 8 和 9,那么就可以把8更新到d[5], 9更新到d[6],得出LIS的长度为6。 然后应该发现一件事情了:在B中插入数据是有序的,而且是进行替换而不需要挪动——也就是说,我们可以使用二分查找,将每一个数字的插入时间优化到O(logN)~~~~~于是算法的时间复杂度就降低到了O(NlogN)~!