最长递增子序列（非连续）Longest Increasing Subsequence

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原文链接：https://my.oschina.net/liyurong/blog/1592166

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#python #数据结构与算法

本文介绍了求解最长递增子序列(LIS)问题的两种高效算法：动态规划与二分查找法。动态规划实现简单，时间复杂度O(n^2)；二分查找法更高效，时间复杂度为O(n log n)。

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2019独角兽企业重金招聘Python工程师标准>>>

问题：

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?

解决：

【题意】求最长递增子序列（可以不连续）

① 使用动态规划。dp[i]表示以nums[i]为结尾的最长递增子串的长度。dp[i] = max(dp[i], dp[j] + 1)，时间复杂度O(n^2)。

class Solution { //32ms
public int lengthOfLIS(int[] nums) {
int[] dp = new int[nums.length];
Arrays.fill(dp,1);
int max = 0;
for (int i = 0;i < nums.length;i ++){
for (int j = 0;j < i;j ++){
if (nums[i] > nums[j]){
dp[i] = Math.max(dp[i],dp[j] + 1);
}
}
max = Math.max(max,dp[i]);
}
return max;
}
}

② 二分查找法，时间复杂度O(nlogn)。使用如下算法：

对于每个数字
     如果（则为list.size（）== 0）
加入列表
      else if（num> list中的最后一个元素）
加入列表
      其他
           替换列表中最小但大于num的元素

class Solution { //3ms
public int lengthOfLIS(int[] nums) {
if (nums == null || nums.length == 0){
return 0;
}
List<Integer> list = new ArrayList<>();
for (int n : nums){
if (list.size() == 0 || n > list.get(list.size() - 1)){
list.add(n);
}else{
int left = 0;
int right = list.size() - 1;
while(left < right){
int mid = (right - left) / 2 + left;
if (list.get(mid) < n){
left = mid + 1;
}else{
right = mid;
}
}
list.set(right,n);
}
}
return list.size();
}
}

③ 将二分搜索的方法提取出来。

class Solution { //0ms
    public int lengthOfLIS(int[] nums) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        int[] record = new int[nums.length + 1];
        record[0] = nums[0];
        int len = 1;
        for(int i = 1; i < nums.length; i++){
            int crt = nums[i];
            if(crt <= record[0]){
                record[0] = crt;
            }else if(crt > record[len - 1]){
                record[len] = crt;
                len ++;
            }else{
                insertNum(record, len, crt);
            }
        }
        return len;
    }

    private void insertNum(int[] record, int len, int crt){
        int start = 0, end = len - 1;
        while(start + 1 < end){
            int mid = start + (end - start) / 2;
            if(record[mid] == crt){
                return;
            }else if(record[mid] < crt){
                start = mid;
            }else{
                end = mid;
            }
        }
        if(record[start] >= crt){
            record[start] = crt;
        }else{
            record[end] = crt;
        }
    }
}

转载于:https://my.oschina.net/liyurong/blog/1592166