动态规划之三：最长上升子序列与最长公共子序列问题

一.最长上升子序列

1.定义

LIS(i):表示以第i个数字为结尾的最长上升子序列的长度

LIS(i):表示在[0...i]的范围内，选择数字nums[i]可以获得的最长上升子序列的长度

LIS(i) = max(1 + LIS(j)) (i > j)

2.例题

300. Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

Input: [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. 

Note:

• 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.

class Solution {
public:
    int lengthOfLIS(vector& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        vector memo(n+1, 1);
        for(int i = 1; i < n; i++){
            for(int j = 0; j < i; j++){
                if(nums[i] > nums[j]){
                    memo[i] = max(memo[i], 1 + memo[j]);
                }
            }
        }
        int res = memo[0];
        for(int i = 1; i < n; i++){
            res = max(res, memo[i]);
        }
        return res;
    }
};

二.最长公共子序列问题

1.定义

给出两个字符串S1和S2，求这两个字符串的最长公共子序列的长度

LCS(m,n) 是 S1[0...m] 和 S2[0...n]的最长公共子序列的长度

S1[m] == S2[n]:  LCS(m, n) = 1 + LCS(m-1, n-1)

S1[m] != S2[n]:  LCS(m, n) = max(LCS(m-1, n), LCS(m, n-1))

#include 
#include 
#include 
#include 
using namespace std;

class Solution {
public:
    int canPartition(vector& nums1,vector& nums2) {
        int m = nums1.size();
        int n = nums2.size();
        vector > memo(m, vector(n, 0));
        if(nums1[0] == nums2[0]) memo[0][0] = 1;
        for(int i=1; i w;
    w.push_back(1);
    w.push_back(2);
    w.push_back(3);
    vector w2;
    w2.push_back(1);
    w2.push_back(2);
    w2.push_back(3);
    w2.push_back(4);
    w2.push_back(2);
    w2.push_back(3);
    int res = solution.canPartition(w, w2);
    printf("%d\n",res);
    return 0;
}