Dynamic programming (the longest increasing subsequence) --- the longest increasing subsequence

The longest increasing subsequence

300. Longest Increasing Subsequence (Medium)

Subject description:

  Given an array, find its longest increasing subsequence

Analysis of ideas:

  Dynamic programming, to define a storage array dp longest increasing subsequence length, dp [n-] represented by the end of the longest increasing sequence length sequence Sn. For an increasing subsequence {Si1, Si2, ..., Sim}, if im <n and Sim <Sn, In this case {Si1, Si2, ..., Sim, Sn} is an increasing sequence, increasing subsequence an increase in length. Incrementing sequence satisfying the above condition, the length of the longest increasing subsequence that is to find, together constitute a Sn containing Sn as the end of the longest increasing the length of the longest sequence in the sequence is incremented. Therefore dp [n] = max {dp [i] +1 | Si <Sn && i <n}.

Code:

public int lengthOfLIS(int []nums){
    if(nums==null||nums.length==0)
        return 0;
    int []dp=new int [nums.length];
    for(int i=0;i<nums.length;i++){
        int max=1;
        for(int j=0;j<i;j++){
            if(nums[i]>nums[j]){
                max=Math.max(max,dp[j]+1);
            }
        }
        dp[i]=max;
    }
    int res=dp[0];
    for(int i=1;i<nums.length;i++){
        if(dp[i]>res)
            res=dp[i];
    }
    return res;
}