LeetCode 300. Longest Increasing Subsequence-优快云博客

本文探讨了如何在未排序整数数组中找到最长递增子序列的长度，并提供了两种算法实现：一种时间复杂度为O(n^2)，另一种通过使用二分查找改进到O(n log n)。同时给出了打印最长递增子序列的方法。

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 is4. 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?

nlgn ---- no idea now how to solve it.

#include <vector>
#include <iostream>
#include <climits>
#include <algorithm>
using namespace std;

// longest increasing sequence.
// dp[i] = max(dp[i], dp[j] + 1) if(a[j] < a[i]), i >= 0 && i <= nums.size(); j >= 0 && j < i;
int lengthOfLIS(vector<int>& nums) {
    if(nums.size() <= 1) return nums.size();
    vector<int> dp(nums.size() + 1, 1);

    for(int i = 1; i <= nums.size(); ++i) {
        for(int j = 0; j < i; ++j) {
            if(nums[j] < nums[i]) {
                dp[i] = max(dp[j] + 1, dp[i]);
            }
        }
    }
    int maxLen = INT_MIN;
    for(int i = 0; i < nums.size(); ++i) {
        maxLen = max(maxLen, dp[i]);
    }
    return maxLen;
}

int main(void) {
    vector<int> nums{10, 9, 2, 5, 3, 7, 101, 18};
    int length = lengthOfLIS(nums);
    cout << length << endl;
}

To print out the longest increasing sequence.

  // print out the increasing sequence
  for(int i = n - 1; i >= 0; --i) {
    if(dp[i] == tmp) {
      cout << nums[i] << endl;
      tmp--;
    }
  }

NLgN method: using binary search

int binarySearch(vector<int>&nums, int left, int right, int target) {
  while(right - left > 1) {
    int mid = left + (right - left) / 2;
    if(nums[mid] >= target) right = mid;
    else left = mid;
  }
  return right;
}

// find the longest increasing subsequence.
int longestIncreasing(vector<int>& nums) {
  int n = nums.size();
  vector<int> res(n, 0);
  int length = 1;
  res.push_back(nums[0]);
  for(int i = 1; i < nums.size(); ++i) {
    if(nums[i] < res[0]) res[0] = nums[i];
    else if(nums[i] > res[length - 1]) res[length++] = nums[i];
    else {
      int index = binarySearch(res, -1, length - 1, nums[i]);
      res[index] = nums[i];
    }
  }
  return length;
}