leetcode全解

1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

  

class Solution {
public:
    vector twoSum(vector& nums, int target) {
        unordered_map num_index_dict;
        vector res;
        for (int i = 0; i < nums.size(); i++) {
            if (num_index_dict.find(nums[i]) != num_index_dict.end()) {
                int index = num_index_dict[nums[i]];
                res.push_back(index);
                res.push_back(i);
                return res;
            } else {
                num_index_dict[target - nums[i]] = i;
            }
        }
        return res;
    }
};

 

2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

  

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry = 0;
        ListNode* dummy_head = new ListNode(0);
        ListNode* head = dummy_head;
        while (l1 || l2 || carry) {
            int temp = carry;
            if (l1) {
                temp += l1 -> val;
                l1 = l1 -> next;
            }
            if (l2) {
                temp += l2 -> val;
                l2 = l2 -> next;
            }
            carry = temp / 10;
            int cur_val = temp % 10;
            head -> next = new ListNode(cur_val);
            head = head -> next;
        }
        return dummy_head -> next;
    }
};

 

3. Longest Substring Without Repeating Characters

 

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

 

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if (s.empty()) {
            return 0;
        }
        int left = 0;
        int right = 0;
        int pos_dict[128];
        for (int i = 0; i < 128; i++) {
            pos_dict[i] = -1;
        }
        int res = 0;
        while (right < s.size()) {
            char cur_char = s[right];
            if (pos_dict[(int)cur_char] == -1) {
                pos_dict[(int)cur_char] = right;
                right += 1;
            } else {
                res = max(res, right - left);
                int last_index = pos_dict[(int)cur_char];
                // 注意此处
                for (int i = left; i<= last_index; i++) {
                    pos_dict[(int)s[i]] = -1;
                }
                left = last_index + 1;
                pos_dict[(int)cur_char] = right;
                right += 1;
            }
        }
        res = max(res, right - left);
        return res;
    }
};

  

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if (s.empty()) {
            return 0;
        }
        int left = 0;
        int right = 0;
        int pos_dict[128];
        for (int i = 0; i < 128; i++) {
            pos_dict[i] = -1;
        }
        int res = 0;
        while (right < s.size()) {
            char cur_char = s[right];
            if (pos_dict[(int)cur_char] == -1) {
                pos_dict[(int)cur_char] = right;
                right += 1;
            } else {
                res = max(res, right - left);
                int last_index = pos_dict[(int)cur_char];
                // 注意此处
                left = max(left, last_index + 1);
                pos_dict[(int)cur_char] = right;
                right += 1;
            }
        }
        res = max(res, right - left);
        return res;
    }
};

 

4. Median of Two Sorted Arrays

 

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0
Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

 

class Solution {
public:
    double findMedianSortedArrays(vector& nums1, vector& nums2) {
        if (nums1.empty() && nums2.empty()) {
            return 0;
        }
        int total_size = nums1.size() + nums2.size();
        vector::iterator iter1 = nums1.begin();
        vector::iterator iter2 = nums2.begin();
        int num = 0;
        int lower;
        int upper;
        while (num < (total_size + 1) / 2) {
            if (iter1 == nums1.end()) {
                lower = *iter2++;
            } else if (iter2 == nums2.end()) {
                lower = *iter1++;
            } else {
                lower = (*iter1 < *iter2) ? *iter1++ : *iter2++;
            }
            num++;
        }
        if (total_size % 2 == 1) {
            return lower;
        }
        if (iter1 == nums1.end()) {
            upper = *iter2;
        } else if (iter2 == nums2.end()) {
            upper = *iter1;
        } else {
            upper = (*iter1 < *iter2) ? *iter1 : *iter2;
        }
        return (lower + upper) / 2.0;
    }
};

 

5. Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example 2:

Input: "cbbd"
Output: "bb"

  

class Solution {
public:
    string longestPalindrome(string s) {
        if (s.size() < 2) {
            return s;
        }
        int arr[2] = {0, 0};
        
        for (int i = 0; i < s.size() - 1; i++) {
            longestPalindromeHelper(s, i, i, arr);
            longestPalindromeHelper(s, i, i + 1, arr);
        }
        return s.substr(arr[0], arr[1]);
    }
    void longestPalindromeHelper(string s, int i, int j, int arr[]) {
        while (i >= 0 && j < s.size() && s[i] == s[j]) {
            i--;
            j++;
        }
        if (j - i - 1 > arr[1]) {
            arr[1] = j - i - 1;
            arr[0] = i + 1;
        }
    }
};

 

class Solution {
public:
    string longestPalindrome(string s) {
        if (s.size() < 2) {
            return s;
        }
        int start = 0;
        int max_length = 1;
        bool dp[s.size()][s.size()] = {false};
        for (int d = 1; d < s.size(); d++) {
            // 注意此处
            for (int i = 0; i + d < s.size(); i++) {
                int j = i + d;
                if (s[i] == s[j] && (d <= 2 || dp[i + 1][j - 1])) {
                    dp[i][j] = true;
                    if (d + 1 > max_length) {
                        max_length = d + 1;
                        start = i;
                    }
                }
            }
        }
        return s.substr(start, max_length);
    }
};

  

6. ZigZag Conversion

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R
And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);
Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

  

class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) {
            return s;
        }
        vector rows(min(int(s.size()), numRows));
        int curRow = 0;
        bool going_down = false;
        for (char c : s) {
            rows[curRow] += c;
            if (curRow == 0 || curRow  == numRows - 1) {
                going_down = !going_down;
            }
            curRow = going_down ? curRow + 1 : curRow - 1;
        }
        string res;
        for (string str : rows) {
            res += str;
        }
        return res;
    }
};

  

class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows  == 1) {
            return s;
        }
        string res = "";
        int cycle_length = 2 * numRows - 2;
        for (int i = 0; i < numRows; i++) {
            for (int j = 0; j + i < s.size(); j += cycle_length) {
                res += s[j + i];
                if (i != 0 && i != numRows - 1 && j + cycle_length - i < s.size()) {
                    res += s[j + cycle_length - i];
                }
            }
        }
        return res;
    }
};

  

 

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