[LeetCode][300. Longest Increasing Subsequence] 3 Approaches: Backtrack, DP, Binary Search

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By Long Luo

This article is the solution 3 Approaches: Backtrack, DP, Binary Search of Problem 300. Longest Increasing Subsequence.

Here shows 3 Approaches to slove this problem: Backtrack, DP, Binary Search.

Backtrack

The Brute Force approach, there are \(2^n\) subsequences, use Backtrack to find the answer.

TLE!

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class Solution {
    int minCount = Integer.MAX_VALUE;

    public int coinChange(int[] coins, int amount) {
        if (amount <= 0) {
            return 0;
        }

        minCount = Integer.MAX_VALUE;
        dfs(coins, amount, 0);
        return minCount == Integer.MAX_VALUE ? -1 : minCount;
    }

    private int dfs(int[] coins, int remain, int count) {
        if (remain < 0) {
            return -1;
        }

        if (remain == 0) {
            minCount = Math.min(minCount, count);
            return count;
        }
        
        for (int x : coins) {
            dfs(coins, remain - x, count + 1);
        }

        return -1;
    }
}

Analysis

  • Time Complexity: \(O(2^n)\)
  • Space Complexity: \(O(n)\)

DP

The equation:

\(dp[i] = \textit{max}(dp[j]) + 1, 0 \leq j < i and \textit{nums}[j] < \textit{nums}[i]\).

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class Solution {
    public int lengthOfLIS(int[] nums) {
        int len = nums.length;
        if (len <= 1) {
            return len;
        }

        int max = 1;
        int[] dp = new int[len];
        Arrays.fill(dp, 1);
        
        for (int i = 1; i < len; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }

            max = Math.max(max, dp[i]);
        }

        return max;
    }
}

Analysis

  • Time Complexity: \(O(n^2)\)
  • Space Complexity: \(O(n)\)
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class Solution {
    public int lengthOfLIS(int[] nums) {
        int len = nums.length;
        if (len <= 1) {
            return len;
        }

        int[] tail = new int[len];
        tail[0] = nums[0];
        int maxLen = 1;

        for (int i = 1; i < len; i++) {
            if (nums[i] > tail[maxLen - 1]) {
                tail[maxLen] = nums[i];
                maxLen++;
            } else {
                int left = 0;
                int right = maxLen - 1;
                while (left < right) {
                    int mid = left + (right - left) / 2;
                    if (tail[mid] < nums[i]) {
                        left = mid + 1;
                    } else {
                        right = mid;
                    }
                }

                tail[left] = nums[i];
            }
        }

        return maxLen;
    }
}

Analysis

  • Time Complexity: \(O(n \log n)\)
  • Space Complexity: \(O(n)\)

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