Edit distance(dynamic programing)

题目描述

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)You have the following 3 operations permitted on a word:a) Insert a character
b) Delete a character
c) Replace a character

题目解析

代码实现

int min3(int a, int b, int c)
{
    return (a < b) ? (a < c ? a : c) : (b < c ? b : c);
}
int minDistance(string word1, string word2) {
    //准备工作：获取字符串长度
    int strlen1 = word1.size();
    int strlen2 = word2.size();
    //准备工作：新建dp矩阵
    vector<vector<int> >dp(strlen1 + 1, vector<int>(strlen2 + 1, 0));
    //第一步：第一行，第一列分别置i
    for (int i = 1; i <= strlen1; i++)
        dp[i][0] = i;
    for (int i = 1; i <= strlen2; i++)
        dp[0][i] = i;
    //第二步：双层for循环进行递推
    for (int i = 1; i <= strlen1; i++)
    {
        for (int j = 1; j <= strlen2; j++)
        {
            int diff;
            if (word1[i - 1] == word2[j - 1])
                diff = 0;
            else
                diff = 1;
            dp[i][j] = min3(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + diff);

        }
    }
    return dp[strlen1][strlen2];
}

Add two numbers

题目描述

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

题目分析

链表版的两个数相加的题，题目是从个位到十位到百位……因此还是比较简单。

代码实现

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    if (l1 == NULL)
        return l2;
    if (l2 == NULL)
        return l1;
    ListNode *result=new ListNode(-1);
    ListNode *p1 = l1;
    ListNode *p2 = l2;
    ListNode *p = result;
    int v = 0;
    int jin = 0;
    while (p1 != NULL || p2 != NULL||jin)
    {
        v = (p1 ? p1->val : 0) + (p2 ? p2->val : 0) + jin;
        jin = v / 10;
        v %= 10;
        ListNode *newnode = new ListNode(v);
        p->next = newnode;
        p = p->next;
        if (p1)
            p1 = p1->next;
        if (p2)
            p2 = p2->next;
    }
    return result->next;
}

partition list

题目描述

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.You should preserve the original relative order of the nodes in each of the two partitions.For example,
Given1->4->3->2->5->2and x = 3,
return1->2->2->4->3->5.

题目解析

最简单的方法，新建两个链表l1和l2，l1存放小于指定数字的链表部分，l2存放大于等于指定数字的链表部分。然后把他们拼接起来。

代码实现

ListNode *partition(ListNode *head, int x)
{
    ListNode *plist1 = new ListNode(0);
    ListNode *plist2 = new ListNode(0);
    ListNode *list1 = plist1;
    ListNode *list2 = plist2;
    while (head != NULL)
    {
        if (head->val < x)
        {
            ListNode *newnode = new ListNode(0);
            newnode->val = head->val;
            newnode->next = NULL;
            list1->next = newnode;
            list1 = list1->next;
        }
        else if (head->val >= x)
        {
            ListNode *bignode = new ListNode(0);
            bignode->val = head->val;
            bignode->next = NULL;
            list2->next = bignode;
            list2 = list2->next;
        }
        head = head->next;
    }
    list1->next = plist2->next;
    plist2->next = NULL;
    return plist1->next;
}