LintCode 76: Longest Increasing Subsequence (最典型的DP题)

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解法1：DP。时间复杂度O(n^2)。代码如下：

class Solution {
public:
    /**
     * @param nums: An integer array
     * @return: The length of LIS (longest increasing subsequence)
     */
    int longestIncreasingSubsequence(vector<int> &nums) {
        int numSize = nums.size();
        if (numSize <= 1) return numSize;
        
        vector<int> DP(numSize, 0);
        int g_maxLen = 0;
        int g_maxId = 0;
    
        DP[0] = 1;
        for (int i = 1; i < numSize; ++i) {
            int maxLen = 0;
            int maxId = 0;
            
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    if (DP[j] > maxLen) {
                        maxLen = DP[j];
                        maxId = j;
                    }
                }
            }
            
            DP[i] = maxLen + 1;
        
            if (DP[i] > g_maxLen) {
                g_maxLen = DP[i];
                g_maxId = i;
            }
        }
        
        return g_maxLen;
    }
};

解法2：Binary Search。 时间复杂度O(nlogn)。
使用一个新的array B[n + 1].
For example, A = {2,3,1}
B = {-INF, INF, INF, INF}

遍历A的每一个元素 a，找到B的第一个大于它的位置，并将该位置的值设为a。例如,
B = {-INF, 2, INF, INF}
B = {-INF, 2, 3, INF}
B = {-INF, 1, 3, INF}

最后，从后往前遍历B的元素，找到第一个非INF的值，其位置即为LIS的值。例如，B[2] = 3, then LIS = 2.

代码如下(参考自九章)：

#include <iostream>
#include <vector>
#include<limits.h>

using namespace std;

// find the first number > num
int binarySearch(vector<int> & minLast, int num) {
    int start = 0, end = minLast.size() - 1;
    while (start + 1 < end) {
        int mid = (end - start) / 2 + start;
        if (minLast[mid] < num) {
            start = mid;
        } else {
            end = mid;
        }
    }

    return end;
}

int longestIncreasingSubsequence(vector<int> & nums) {
    vector<int> minLast(nums.size() + 1, 0);
    minLast[0] = INT_MIN;

    for (int i = 1; i <= nums.size(); i++) {
        minLast[i] = INT_MAX;
    }

    for (int i = 0; i < nums.size(); i++) {
        // find the first number in minLast >= nums[i]
        int index = binarySearch(minLast, nums[i]);
        minLast[index] = nums[i];
    }

    for (int i = nums.size(); i >= 1; i--) {
        if (minLast[i] != INT_MAX) {
            return i;
        }
    }

    return 0;
}

int main()
{
  //  vector<int> nums = {2, 3, 1, 7, 6, 4, 8};
    vector<int> nums = {2, 3, 1};
    int result;
    result = longestIncreasingSubsequence(nums);
    cout<<result<<endl;
}