题目

给定一个无序整型数组, 找出最大的递增子序列的长度.

Input: [10, 9, 2, 5, 3, 7, 101, 18]
Output: 4  // [2,3,7,101]

思路1

递归.

int lengthOfLISHelper(vector<int>& nums, int flag, int pos) {
    if (pos >= nums.size()) return 0;
    int count = 0;
    if (nums[pos] > flag) {
        count = lengthOfLISHelper(nums, nums[pos], pos+1) + 1;
    }
    int nextCount = lengthOfLISHelper(nums, flag, pos+1);
    return max(nextCount, count);
}

int lengthOfLIS(vector<int>& nums) {
    return lengthOfLISHelper(nums, INT_MIN, 0);
}

思路2

DP.

int lengthOfLIS2(vector<int>& nums) {
    if (nums.empty()) return 0;
    // 记录nums[i]的最大子序列个数
    vector<int> dp(nums.size());
    int count = 1;
    dp[0] = 1;

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

总结

求最值, 优先考虑使用DP. 熟练掌握递归思想, 递归难以理解, 但是可以熟能生巧.