标签：... s2 s1 牛客 word1 word2 机试 dp HJ52

原题传送门

1. 问题描述

2. Solution

java版思路分析，来自LeetCode 72

/*
    DP
    1. Define the state dp[i][j]
        to be the minimum number of operations to convert word1[0...i) to word2[0...j)
    2. Initial state
        For the base case, that is, to convert a string to an empty string, the minimum number of operations(deletions)
        is just the length of the string. So we have dp[i][0] = i and dp[0][j] = j.
    3. State transition equation
        For the general case to convert word1[0...i) to word2[0...j), we break this problem down into sub-problems.
        Suppose we have already known how to convert word1[0...i-1) to word2[0...j-1) i.e. dp[i-1][j-1],
        if word1[i-1] == word2[j-1], then no more operation is needed and dp[i][j] = dp[i-1][j-1]

        if word1[i-1] != word2[j-1], we need to consider three cases.
        a) Replace word1[i-1] by word2[j-1] i.e. dp[i][j] = dp[i-1][j-1] + 1
        b) if word1[0...i-1) = word2[0...j) then delete word1[i-1] i.e. dp[i][j] = dp[i-1][j] + 1
        c) if word1[0...i) + word2[j-1) = word2[0...j) then insert word2[j-1] to word1[0...i)
            i.e. dp[i][j] = dp[i][j-1] + 1
    So when word1[i-1] != word2[j-1], dp[i][j] will just be the minimum of the above three case
*/
public class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) dp[i][0] = i;
        for (int j = 1; j <= n; j++) dp[0][j] = j;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1))
                    dp[i][j] = dp[i - 1][j - 1];
                else
                    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
            }
        }
        return dp[m][n];
    }
}

本题题解

import sys

if sys.platform != "linux":
    file_in = open("input/HJ52.txt")
    sys.stdin = file_in

"""
DP

定义: dp[i][j] = s1[i] 与 s2[j] 之间的距离
初始化: 
    对于边界，是转换一个string到空string，使用删除，所以dp值为原始string的长度
    dp[i][0] = i, dp[0][j] = j
状态转移方程:

if s1[i] == s2[j] 
    dp[i][j] = dp[i-1][j-1]
if s1[i] != s2[j]
    a) 替换s2[j] -> s1[i]: dp[i][j] = dp[i - 1][j-1] + 1
    b) 删除s1[i]:  dp[i][j] = dp[i-1][j] + 1
    c) 插入 s1[0...i] + s2[j-1] = s2[0...j]: dp[i][j] = dp[i][j-1] + 1
    综上，dp[i][j] = min{dp[i-1][j-1], dp[i-1][j], dp[i][j-1]} + 1
"""


def solve(s1, s2):
    m = len(s1)
    n = len(s2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if s1[i - 1] == s2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
    print(dp[m][n])


while True:
    try:
        s1 = input().strip()
        s2 = input().strip()
        solve(s1, s2)
    except:
        break

标签：...,s2,s1,牛客,word1,word2,机试,dp,HJ52
来源： https://www.cnblogs.com/junstat/p/16172574.html

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