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?
思路:n2的复杂度,用一个dp存每个元素作为末尾元素的最长序列长度
public class Solution {
public int lengthOfLIS(int[] nums) {
if(nums.length==0){
return 0;
}
int m=1;
int dp[]=new int[nums.length];
dp[0]=1;
for(int i=1;i<nums.length;i++){
int max=0;
for(int j=0;j<i;j++){
if(nums[j]<nums[i]){
max=Math.max(dp[j],max);
}
}
dp[i]=max+1;
m=Math.max(m,dp[i]);
}
return m;
}
}