# 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

## 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] = i; for (int i = 1; i <= strlen2; i++) dp[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];}``

## 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

## 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.

## 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;}``