300. Longest Increasing Subsequence【力扣】

题意理解

给定一个字符串,问最长的递增子序列有多长,(可以不连续)。

问题分析

用动规,状态量,状态转移方程

状态量 DP[i] 前i个字符(带第i个字符)的递增子序列最长长度

状态转移方程

DP[i] = max( DP[j] ) + 1,   0 <= j <= i;

这次的规律不是i个状态和i-1个状态的关系,而是i个状态和0..i-1个状态的关系。

最终的最大值也是所有最大值的最大值。

 

其他

方法是这样,但道理有点麻烦,没弄通。

This method relies on the fact that the longest increasing subsequence possible upto the i^{th}ith index in a given array is independent of the elements coming later on in the array. Thus, if we know the length of the LIS upto i^{th}ith index, we can figure out the length of the LIS possible by including the (i+1)^{th}(i+1)th element based on the elements with indices jj such that 0 \leq j \leq (i + 1)0≤j≤(i+1).

链接

    int lengthOfLIS(vector& nums) {
        if (nums.size() == 0) 
            return 0;
        if (nums.size() == 1)
            return 1;
        int dp[nums.size()];
        dp[0] = 1;
        for (int i = 1; i < nums.size(); i++)
        {
            int maxVal = 0;
            for(int j = 0; j < i; j ++)
            {
                if (nums[j] < nums[i])
                {
                    maxVal = max(maxVal, dp[j]);
                }
            }
            dp[i] = maxVal + 1;
            cout << i << '\t' << dp[i] << endl;
        }
        int maxLen = 0;
        for(int i = 0; i < nums.size(); i++)
            maxLen = max(maxLen, dp[i]);
        return maxLen;
    }

 

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