LeetCode 300. Longest Increasing Subsequence

捭阖寰宇

于 2016-09-07 11:32:39 发布

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分类专栏：算法

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本文介绍了一种求解最长递增子序列长度的算法，给出了一段Java代码实现，并指出算法的时间复杂度为O(n^2)。该问题旨在找出数组中最长递增子序列的长度。

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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.

注意：该问题求的是递增子序列而不是递增子串

Java代码如下所示：

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums == null || nums.length == 0) {
        	return 0;
        }
        int max = 1;
        int[] dp = new int[nums.length];
        dp[0] = 1;
        for(int i = 1; i < dp.length; i++) {
        	int preMax = 0;
        	for(int j = 0; j < i; j++) {
        		if(nums[j] < nums[i] && preMax < dp[j]) {
        			preMax = dp[j];
        		}
        	}
        	dp[i] = preMax + 1;
        	if(max < dp[i]) {
        		max = dp[i];
        	}
        }
        return max;
    }
    
    public static void main(String[] args) {
    	Solution solution = new Solution();
		int[] nums = {10, 9, 2, 5, 3, 7, 101, 18};
		System.out.println(solution.lengthOfLIS(nums));
	}
}empty