lintcode:最长上升子序列

题目

最长上升子序列

给定一个整数序列，找到最长上升子序列（LIS），返回LIS的长度。

样例

给出[5,4,1,2,3]，这个LIS是[1,2,3]，返回 3

给出[4,2,4,5,3,7]，这个LIS是[4,4,5,7]，返回 4

挑战

要求时间复杂度为O(n^2) 或者O(nlogn)

说明

最长上升子序列的定义：

最长上升子序列问题是在一个无序的给定序列中找到一个尽可能长的由低到高排列的子序列，这种子序列不一定是连续的或者唯一的。
https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
解题

下面是很久前复制网上的程序

public class Solution {
/**
* @param nums: The integer array
* @return: The length of LIS (longest increasing subsequence)
*/
public int longestIncreasingSubsequence(int[] nums) {
int len = nums.length;
if(nums == null || len ==0)
return 0;
ArrayList<Integer> dp = new ArrayList<Integer>();
for(int i=0;i<len ;i++){
if(dp.isEmpty() || dp.get(dp.size() - 1) <= nums[i])
dp.add(nums[i]);
else{
int index = findFirstLargeEqual(dp,nums[i]);
dp.set(index,nums[i]);

}
}
return dp.size();
}
public int findFirstLargeEqual(ArrayList<Integer> list,int num){
int left = 0;
int right = list.size() - 1;
while(left < right){
int mid = (left + right)/2;
if(list.get(mid) <= num)
left = mid + 1;
else
right = mid;
}
return left;
}
}

Java Code
Python

class Solution:
"""
@param nums: The integer array
@return: The length of LIS (longest increasing subsequence)
"""
def longestIncreasingSubsequence(self, nums):
# write your code here
if nums == None or len(nums) ==0:
return 0
lst = list()
for i in range(len(nums)):
if len(lst) == 0 or lst[len(lst) - 1] <= nums[i]:
lst.append(nums[i])
else:
index = self.findFirstLargeEqual(lst,nums[i])
lst[index] = nums[i]
return len(lst)

def findFirstLargeEqual(self,lst,target):
left = 0
right = len(lst) -1
while left < right:
mid = (left + right)/2
if lst[mid] <= target:
left = mid + 1
else:
right = mid
return left