300. Longest Increasing Subsequence (M)

Longest Increasing Subsequence (M)

Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4 
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. 

Note:

• There may be more than one LIS combination, it is only necessary for you to return the length.
• Your algorithm should run in O($n^2$) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

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题意

找到一个无序数组中的最长递增子序列。

思路

1. 暴力法，直接使用dfs搜索；
2. 对暴力法进行优化，将每一次dfs的结果记录下来；
3. 动态规划。dp[i]代表以nums[i]为结尾的最长递增子数列的长度。针对一个i，遍历 j (j = 0, 1, …, i - 1)，如果nums[i]>nums[j]，说明可以将nums[i]添加到nums[j]后面构成一个新的递增子数列，其长度为dp[j]+1，这时候更新dp[i]为dp[i]和dp[j]+1之中较大的一个即可；
4. 动态规划二分优化。遍历nums数组，针对每一个nums[i]，用二分法查找它在dp数组中应该出现的位置x，如果该位置上已经有数，则更新dp[x]=nums[i]；如果该位置上还没有数，则将nums[i]插入到该位置。

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代码实现 - dfs暴力

class Solution {
    public int lengthOfLIS(int[] nums) {
        return dfs(nums, -1, 0);
    }

    private int dfs(int[] nums, int pre, int cur) {
        if (cur == nums.length) {
            return 0;
        }

        int use = (pre < 0 || nums[cur] > nums[pre]) ? 1 + dfs(nums, cur, cur + 1) : 0;
        int skip = dfs(nums, pre, cur + 1);
        return Math.max(use, skip);
    }
}

代码实现 - dfs优化

class Solution {
    public int lengthOfLIS(int[] nums) {
        int[][] memo = new int[nums.length][nums.length];
        for (int[] list : memo) {
            Arrays.fill(list, -1);
        }
        return dfs(nums, -1, 0, memo);
    }

    private int dfs(int[] nums, int pre, int cur, int[][] memo) {
        if (cur == nums.length) {
            return 0;
        }
        if (memo[pre + 1][cur] >= 0) {
            return memo[pre + 1][cur];
        }

        int use = (pre < 0 || nums[cur] > nums[pre]) ? 1 + dfs(nums, cur, cur + 1, memo) : 0;
        int skip = dfs(nums, pre, cur + 1, memo);
        memo[pre + 1][cur] = Math.max(use, skip);
        return memo[pre + 1][cur];
    }
}

代码实现 - 动态规划

class Solution {
    public int lengthOfLIS(int[] nums) {
        int[] dp = new int[nums.length];
        int maxLen = 0;

        for (int i = 0; i < nums.length; i++) {
            dp[i] = 1;		// 初始值为1
            for (int j = 0; j < i; j++) {
                dp[i] = nums[i] > nums[j] ? Math.max(dp[i], dp[j] + 1) : dp[i];
            }
            maxLen = Math.max(maxLen, dp[i]);
        }
        return maxLen;
    }
}

代码实现 - 动态规划二分优化

class Solution {
    public int lengthOfLIS(int[] nums) {
        List<Integer> dp = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            int pos = binarySearch(dp, nums[i]);
            if (pos == dp.size()) {
                dp.add(nums[i]);
            } else {
                dp.set(pos, nums[i]);
            }
        }
        return dp.size();
    }

    private int binarySearch(List<Integer> list, int target) {
        int left = 0, right = list.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (list.get(mid) < target) {
                left = mid + 1;
            } else if (list.get(mid) > target) {
                right = mid - 1;
            } else {
                return mid;
            }
        }
        return left;
    }
}