300. Longest Increasing Subsequence

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?

My Solution(C/C++)

#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;

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

int main() {
    vector<int> nums;
    nums.push_back(1);
    nums.push_back(3);
    nums.push_back(2);
    nums.push_back(3);
    nums.push_back(1);
    nums.push_back(4);
    Solution s;
    printf("最長上升子序列的長度為：%d", s.lengthOfLIS(nums));
    return 0;
}

結(jié)果

最長上升子序列的長度為：4

My Solution2.0(C/C++)

class Solution {
public:
    int lengthOfLIS(vector<int> &nums) {
        if (nums.size() == 0) {
            return 0;
        }
        vector<int> stack;
        for (int i = 0; i < nums.size(); i++) {
            if (stack.empty() || stack.back() < nums[i]) {
                stack.push_back(nums[i]);
            }
            else {
                for (int j = 0; j < stack.size(); j++) {
                    if (stack[j] >= nums[i]) {  //與棧中元素相等時也要替換，不然棧中可能會同時存入多個相同元素
                        stack[j] = nums[i];
                        break;
                    }
                }
            }
        }
        return stack.size();
    }
};

My Solution2.0plus(C/C++, n*logn)

class Solution {
public:
    int lengthOfLIS(vector<int> &nums) {
        if (nums.size() == 0) {
            return 0;
        }
        vector<int> stack;
        for (int i = 0; i < nums.size(); i++) {
            if (stack.empty() || stack.back() < nums[i]) {
                stack.push_back(nums[i]);
            }
            else {
                int k = getKthOfStack(stack, nums[i]);
                stack[k] = nums[i];
            }
        }
        return stack.size();
    }
private:
    int getKthOfStack(vector<int> &stack, int num) {
        int begin = 0;
        int end = stack.size() - 1;
        int mid, index;
        while (begin <= end) {
            mid = (begin + end) / 2;
            if (num == stack[mid]) {
                return mid;
            }
            else if (num < stack[mid]) {
                end = mid - 1;
                if (mid == 0 || num > stack[mid - 1]) {
                    index = mid;
                }
            }
            else if (num > stack[mid]) {
                begin = mid + 1;
                if (mid == stack.size() - 1 || num < stack[mid + 1]) {
                    index = mid + 1;
                }
            }
        }
        return index;
    }

My Solution(Python)

class Solution:
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if nums == []:
            return 0
        max_dp, result = 0, 1
        dp = [[nums[i]] for i in range(len(nums))]
        dp[0].append(1)
        for i in range(1, len(nums)):
            for j in range(i):
                if dp[i][0] > dp[j][0]:
                    max_dp = max(max_dp, dp[j][1])
            dp[i].append(max_dp + 1)
            result = max(result, max_dp + 1)
            max_dp = 0
        return result