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Given an unsorted array of integers, find the length of longest increasing subsequence.
For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that 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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
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题意：无需多说。主要是这玩意有基础版O（n^2）和增强版O(nlogn)
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public class Solution {
public int lengthOfLIS(int[] nums) {
int length=nums.length;
if(length==0)return 0;
int max=1;
int[] d=new int[length];
for(int i=1;i<length;i++){
d[i]=1;
for(int j=0;j<i;j++){
if(nums[j]<nums[i])
d[i]=Math.min(d[i],d[j]=1);
}
max=Math.max(d[i],max);
}
return max;
}
}
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PS:思路参加http://www.hawstein.com/posts/dp-novice-to-advanced.html
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public class AscentSequence {
public int findLongest(int[] A, int n) {
int[] dp=new int[n];
int[] ends=new int[n];
ends[0]=A[0];
dp[0]=1;
int max=1;
int r=0,right=0,l=0,m=0;
for(int i=1;i<n;i++){
l=0;
r=right;
while(l<=r){
m=(l+r)/2;
if(A[i]>ends[m]){
l=m+1;
}else{
r=m-1;
}
}
right=Math.max(l,right);
ends[l]=A[i];
dp[i]=l+1;
max=Math.max(dp[i],max);
}
return max;
}
}
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用二分查找、辅助数组。
最后是打印出来最长递增子序列
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public static int[] getLIS(int[] arr,int[] dp){
int maxlen=0;
int index=0;
for(int i=0;i<dp.length;i++){
if(dp[i]>maxlen){
maxlen=dp[i];
index=i;
}
}
System.out.println(index+""+maxlen);
int[] list=new int[maxlen];
list[--maxlen]=arr[index];
for(int j=index;j>=0;j--){
if(arr[j]<arr[index] && dp[j]+1==dp[index]){
list[--maxlen]=arr[j];
index=j;
}
}
return list;
}
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PS：首先找到dp中最大值及其索引index。然后去数组arr[index]中从右往左遍历。找到第一个小于arr[index]且dp[]+1==dp[index]的值作为倒数第二个数。一次往后重复此过程。
本文转自 努力的C 51CTO博客，原文链接:http://blog.51cto.com/fulin0532/1903095
