leetcode 21-30 easy

21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode dummy(INT_MIN);
        ListNode *tail=&dummy;
        while(l1 && l2)
        {
            if(l1->val<l2->val)
            {
                tail->next=l1;
                l1=l1->next;
                
            }
            else 
            {
                tail->next=l2;
                l2=l2->next;
            }
            tail=tail->next;
        }
        
        //谁没结束就补充谁
        tail->next=l1?l1:l2;
        return dummy.next;    
    }
};

python 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists1(self, l1, l2):
        dummy = cur = ListNode(0)
        while l1 and l2:
            if l1.val < l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        cur.next = l1 or l2
        return dummy.next
    
# recursively    
    def mergeTwoLists2(self, l1, l2):
        if not l1 or not l2:
            return l1 or l2
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2
        
# in-place, iteratively        
    def mergeTwoLists(self, l1, l2):
        if None in (l1, l2):
            return l1 or l2
        dummy = cur = ListNode(0)
        dummy.next = l1
        while l1 and l2:
            if l1.val < l2.val:
                l1 = l1.next
            else:
                nxt = cur.next
                cur.next = l2
                tmp = l2.next
                l2.next = nxt
                l2 = tmp
            cur = cur.next
        cur.next = l1 or l2
        return dummy.next

26. Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

C++

int removeDuplicates(vector<int>& nums) {
    int i = 0;
    for (int n : nums)
        if (!i || n > nums[i-1])
            nums[i++] = n;
    return i;
}

python

from collections import OrderedDict
class Solution(object):
    def removeDuplicates(self, nums):

        nums[:] =  OrderedDict.fromkeys(nums).keys()
        return len(nums)



###########
class Solution:
    # @param a list of integers
    # @return an integer
    def removeDuplicates(self, A):
        if not A:
            return 0
        end = len(A)
        read = 1
        write = 1
        while read < end:
            if A[read] != A[read-1]:
                A[write] = A[read]
                write += 1
            read += 1
        return write

28、Implement strStr()

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().

C++

//暴力算法
class Solution {
public:
    int strStr(string haystack, string needle) {
        int m = haystack.length(), n = needle.length();
        if (!n) return 0;
       
        for (int i = 0; i < m - n + 1; i++) {
            int j = 0;
            for (; j < n; j++) {
                if (haystack[i + j] != needle[j]) {
                    break;
                }
            }
            if (j == n)  return i;
        }
        return -1;
    }
};

//KMP

class Solution {
public:
    int strStr(string haystack, string needle) {
        int m = haystack.length(), n = needle.length();
        if (!n) {
            return 0;
        }
        vector<int> lps = kmpProcess(needle);
        for (int i = 0, j = 0; i < m; ) {
            if (haystack[i] == needle[j]) { 
                i++;
                j++;
            }
            if (j == n) {
                return i - j;
            }
            if ((i < m) && (haystack[i] != needle[j])) {
                if (j) {
                    j = lps[j - 1];
                }
                else {
                    i++;
                }
            }
        }
        return -1;
    }
private:
    vector<int> kmpProcess(string& needle) {
        int n = needle.length();
        vector<int> lps(n, 0);
        for (int i = 1, len = 0; i < n; ) {
            if (needle[i] == needle[len]) {
                lps[i++] = ++len;
            } else if (len) {
                len = lps[len - 1];
            } else {
                lps[i++] = 0;
            }
        }
        return lps;
    }
};