题目如下:

解题思路:本题和【leetcode】583. Delete Operation for Two Strings 类似,区别在于word1[i] != word2[j]的时候,是删除word1[i]还是word2[j]取决于min(dp[i-1][j]+ord(word1[i-1]),dp[i][j-1]+ord(word2[j-1]))。

代码如下:

class Solution(object):
def minimumDeleteSum(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: int
"""
word1 = s1
word2 = s2
if len(word1) == 0 or len(word2) == 0:
return abs(len(word2) - len(word1))
dp = [[0 for x in range(len(word2)+1)] for x in range(len(word1)+1)]
w1_ascii = 0
w2_ascii = 0
for i in xrange(1,len(word1)+1):
w1_ascii += ord(word1[i-1])
dp[i][0] = w1_ascii
for j in xrange(1,len(word2)+1):
w2_ascii += ord(word2[j-1])
dp[0][j] = w2_ascii
for i in xrange(1,len(word1)+1):
for j in xrange(1,len(word2)+1):
if word1[i-1] == word2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = min(dp[i-1][j]+ord(word1[i-1]),dp[i][j-1]+ord(word2[j-1]))
return dp[-1][-1]

【leetcode】712. Minimum ASCII Delete Sum for Two Strings的更多相关文章

  1. 【LeetCode】712. Minimum ASCII Delete Sum for Two Strings 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...

  2. LN : leetcode 712 Minimum ASCII Delete Sum for Two Strings

    lc 712 Minimum ASCII Delete Sum for Two Strings 712 Minimum ASCII Delete Sum for Two Strings Given t ...

  3. LC 712. Minimum ASCII Delete Sum for Two Strings

    Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. ...

  4. [LeetCode] 712. Minimum ASCII Delete Sum for Two Strings 两个字符串的最小ASCII删除和

    Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. ...

  5. LeetCode 712. Minimum ASCII Delete Sum for Two Strings

    Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. ...

  6. 712. Minimum ASCII Delete Sum for Two Strings

    题目: Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings eq ...

  7. Leetcode之动态规划(DP)专题-712. 两个字符串的最小ASCII删除和(Minimum ASCII Delete Sum for Two Strings)

    Leetcode之动态规划(DP)专题-712. 两个字符串的最小ASCII删除和(Minimum ASCII Delete Sum for Two Strings) 给定两个字符串s1, s2,找到 ...

  8. [LeetCode] Minimum ASCII Delete Sum for Two Strings 两个字符串的最小ASCII删除和

    Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal. ...

  9. 【LeetCode】931. Minimum Falling Path Sum 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 动态规划 相似题目 参考资料 日期 题目地址:htt ...

随机推荐

  1. join当前线程等待指定的线程结束后才能继续运行

    模拟一个QQ游戏大厅斗地主 /** sleep(休眠.睡眠) join当前线程等待指定的线程结束后才能继续运行 */ class Player extends Thread{ private Stri ...

  2. leetcode 111二叉树的最小深度

    使用深度优先搜索:时间复杂度O(n),空间复杂度O(logn) /** * Definition for a binary tree node. * struct TreeNode { * int v ...

  3. 使用Dockerfile封装Django镜像

    第一步: 在/opt下建立了docker目录,下载一个django-2.1.7的源码包, touch Dockerfile和run.sh,其中run.sh是用来执行Django的bash脚本,Dock ...

  4. import * as 用法

  5. curl发json

    linux 模拟post请求 curl -X POST \ -H "Content-Type: application/json" \ -H "token:GXJP1cl ...

  6. 关于fork

    关于fork 之前和同学讨论了一个关于fork的问题,之前自己也是稍微看过一点,但是具体的也不是太了解,这样还是很不好的. 具体的问题来源于一个面试题,大概是问 fork||fork操作会生成几个新的 ...

  7. linux如何处理多连接请求?

    1.TCP迭代服务器程序 这种方式就是服务器同一时间只处理一个客户端的请求,这个请求处理完以后才转向下一个客户请求.当然这样的服务器程序比较少见,这就像一个公司只能一次处理一个客户,后面的客户只能等待 ...

  8. xmake v2.1.9版本发布,增加可视化图形菜单配置

    此版本主要增加xmake f --menu实现用户自定义图形菜单配置,界面风格类似linux的make menuconfig: [图片上传失败-(image-505bc0-1517795319124) ...

  9. [Python3] 015 冰冻集合的内置方法

    目录 0. 前言 英文名 元素要求 使用限制 返回 方法数量 1. 如何查看 frozenset() 的内置方法 2. 少废话,上例子 2.1 copy() 2.2 difference() 2.3 ...

  10. sqlMap.xml配置文件中迭代一个集合的方式

    比如:根据班级号查询学生的信息,参数是list 1.foreach的用法:[写法一] <select id="getStudentListByClassId" resultM ...