300. Longest Increasing Subsequence
Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
Example 1:
Input: nums = [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.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
- 1 <= nums.length <= 2500
- -104 <= nums[i] <= 104
Follow up:
- Could you come up with the O(n2) solution?
- Could you improve it to O(n log(n)) time complexity?
class Solution {
public int lengthOfLIS(int[] nums) {
int maxLen = 1;
int[] dp = new int[nums.length];
for(int i=0;i<nums.length;i++){
dp[i] = 1;
for(int j=0;j<i;j++){
if(nums[j]<nums[i]){
dp[i] = Math.max(dp[i], dp[j]+1);
}
}
maxLen = Math.max(maxLen,dp[i]);
}
return maxLen;
}
}初始条件： 在最开始时，dp[i] = 1代表序列中只含有自身。
转移方程： 个人本问题的关键在于对于循环和大小关系的确定：我们对于任意的一个i，需要从0～i-1判断每个nums[i],nums[j]的大小关系。从而确定dp[i]的值，通过两者之间的大小关系我们可以得出： 若nums[i] > nums[j],则dp[i] = dp[j]+1 然后对于j从0～i-1，我们对上述式子求最大值，便可得出转移方程如下： dp[i] = max(dp[i],dp[j]+1)