491. Increasing Subsequences 491.增加子序列

Given an integer array, your task is to find all the different possible increasing subsequences of the given array, and the length of an increasing subsequence should be at least 2.

 

Example:

Input: [4, 6, 7, 7]
Output: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]

和划分字符串一样的思路，就是把isPalindrome换成isIncreasing就行了
如果出现了空串、长度为1的串，需要限制一下temp的尺寸大于等于2

class Solution {
    public List> findSubsequences(int[] nums) {
        //cc
        List temp = new ArrayList<>();
        List> result = new ArrayList<>();
        
        if (nums == null) return null;
        if (nums.length == 0) return result;
        
        dfs(nums, 0, temp, result);
        
        return result;
    }
    
    public void dfs(int[] nums, int start, List temp, List> result) {
        //控制一下size
        if (start == nums.length && temp.size() >= 2) {
            result.add(new ArrayList(temp));
        }else {
            for (int i = start; i < nums.length; i++) {
                if (isIncreasing(nums, start, i)) {
                    //temp.add(Arrays.copyOfRange(nums, start, i));
                    for (int j = start; j < i; j++) {
                        temp.add(nums[j]);
                    }
                    dfs(nums, i + 1, temp, result);
                    
                    if (temp.size() >= 1)
                    temp.remove(temp.size() - 1);
                }
            }
        }
    }
    
    //isIncreasing
    public boolean isIncreasing(int[] nums, int start, int end) {
        for (int i = start; i < end - 1; i++) {
            if (nums[i + 1] < nums[i])
                return false;
        }
        return true;
    }
}

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