LeetCode 491. Increasing Subsequences

【题目】

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]]

Note:

1. The length of the given array will not exceed 15.
2. The range of integer in the given array is [-100,100].
3. The given array may contain duplicates, and two equal integers should also be considered as a special case of increasing sequence.

【题解】

这道题用递归的思路，vecTmp存储当前考虑的子序列，nPos表示vecTmp中最后一个元素的下标的后一个(需要注意nPos为0的特殊情况)，则如果nums[nPos]的元素大于或等于vecTmp中的最后一个元素，就可以将其增加到vecTmp中，然后继续处理nPos+1处的元素，直到最后一个元素或者元素比最后一个小。接着返回原始的vecTmp，将nums[nPos]弹出，继续下一次递归。

同时，由于数组的元素是可以重复的，所以需要注意将重复结果去除，这里用容器set来简化这些过程。

【代码】

    vector> findSubsequences(vector& nums) {
        vector> vecResult;
        int nSize = nums.size();
        if (nSize == 0 || nSize == 1)
            return vecResult;
        
        set> setResult;
        GetResult(setResult, nums, vector(), 0, nums.size());
        
        return vector>(setResult.begin(), setResult.end());
    }
    
    void GetResult(set>& setResult, vector nums, vector vecTmp, int nPos, int nSize) {
        if (int(vecTmp.size()) > 1)
            setResult.insert(vecTmp);
        
        for (int i = nPos; i < nSize; i++) {
            int nTmpSize = vecTmp.size();
            if (nTmpSize == 0 || vecTmp[nTmpSize - 1] <= nums[i]) {
                vecTmp.push_back(nums[i]);
                GetResult(setResult, nums, vecTmp, i + 1, nSize);
                int nCurLen = vecTmp.size();
                vecTmp.erase(vecTmp.begin() + nCurLen - 1);
            }
        }
    }