LeetCode - 268. Missing Number & 674. Longest Continuous Increasing Subsequence

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LeetCode - 268. Missing Number & 674. Longest Continuous Increasing Subsequence

• LeetCode - 268. Missing Number
• LeetCode - 674. Longest Continuous Increasing Subsequence

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LeetCode - 268. Missing Number

题目链接

题目

解析

题目要求在O(n)时间和O(1)空间内完成。
也是一个很有意思的题目，简单的想法:

• 用一个变量sumAll记录包含没有丢失的那个数的所有的和(也可以用等差数列求和公式求出)；
• 然后求出数组的和sum，结果就是sumAll - sum；

更好的解法，利用亦或的性质:

利用第四条性质，循环亦或xor = xor ^ (i+1) ^ nums[i]其中没有出现的数就会剩下来。

class Solution {
    public int missingNumber(int[] nums) {
        int sumAll = 0, sum = 0;
        for(int i = 0; i < nums.length; i++){
            sumAll += (i+1);
            sum += nums[i];
        }
        return sumAll-sum;
    }
}

用等差数列求和公式求出sumAll：

class Solution {
    
    public int missingNumber(int[] nums) {
        int sumAll = (0+nums.length)*(nums.length+1)/2; //等差数列
        int sum = 0;
        for(int i = 0; i < nums.length; i++)
            sum += nums[i];
        return sumAll-sum;
    }
    
}

更加巧妙的方式，利用异或运算:

class Solution {

    public int missingNumber(int[] nums) {
        int xor = 0;
        for(int i = 0; i < nums.length; i++)
            xor = xor ^ (i+1) ^ nums[i];
        return xor;
    }
}

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LeetCode - 674. Longest Continuous Increasing Subsequence

题目链接

题目

求最长连续递增子串(不是子序列)。

解析

比最长递增子序列更简单，同样提供三种解法。

一维dp

class Solution {
    
    public int findLengthOfLCIS(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int res = 1;
        int[] dp = new int[nums.length];
        dp[0] = 1;
        for(int i = 1; i < nums.length; i++){
            if(nums[i] > nums[i-1])
                dp[i] = dp[i-1] + 1;
            else 
                dp[i] = 1;
            res = Math.max(res, dp[i]); 
        }
        return res;
    }

}

记忆化递归:

class Solution {   
    
    public int res;
    public int findLengthOfLCIS(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int[] dp = new int[nums.length];
        Arrays.fill(dp, -1);
        res = 1;
        process(nums, nums.length-1, dp);
        return res;
    }
    public int process(int[] nums, int i, int[] dp){
        if(i == 0)
            return 1;
        if(dp[i] != -1)
            return dp[i];
        if(nums[i] > nums[i-1]){
            dp[i] = process(nums, i-1, dp) + 1;
        }else {
            dp[i] = 1;
            process(nums, i-1, dp); // still should 
        }
        res = Math.max(res, dp[i]);
        return dp[i];
    }
}

滚动优化

class Solution {   

    public int findLengthOfLCIS(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int res = 1, cur = 1;
        for(int i = 1; i < nums.length; i++){
            if(nums[i] > nums[i-1])
                cur += 1;
            else 
                cur = 1;
            res = Math.max(res, cur); 
        }
        return res;
    }
}

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