給出兩個(gè)單詞word1和word2劣针，計(jì)算出將word1 轉(zhuǎn)換為word2的最少操作次數(shù)。
你總共三種操作方法：
插入一個(gè)字符
刪除一個(gè)字符
替換一個(gè)字符
樣例
給出 work1="mart" 和 work2="karma"
返回 3
分析:minSteps[i][j]表示word1的前i個(gè)字符改為word2的前j個(gè)字符的最少操作數(shù)，因此有轉(zhuǎn)移方程
minSteps[i][j] =
{
minSteps[i - 1][j - 1];(word1[i] == wrod2[j])
min(minSteps[i - 1][j - 1],minSteps[i][j - 1],minSteps[i - 1][j]);(word1[i] != wrod2[j])
}
public class Solution {
public int minDistance(String word1, String word2) {
if(null == word1 || null == word2)
return 0;
int[][] minSteps = new int[word1.length() + 1][word2.length() + 1];
for(int i = 0;i <= word1.length();i++)
{
minSteps[i][0] = i;
}
for(int i = 0;i <= word2.length();i++)
{
minSteps[0][i] = i;
}
for(int i = 1;i <= word1.length();i++)
{
for(int j = 1;j <= word2.length();j++)
{
if(word1.charAt(i - 1) == word2.charAt(j - 1))
{
minSteps[i][j] = minSteps[i - 1][j - 1];
}
else
{
minSteps[i][j] = Math.min(minSteps[i - 1][j - 1], Math.min(minSteps[i - 1][j], minSteps[i][j - 1])) + 1;
}
}
}
return minSteps[word1.length()][word2.length()];
}
};
