匹配两个字符串并在java中打印缺失的单词答案

【问题标题】：match two strings and print the missing words in java匹配两个字符串并在java中打印缺失的单词
【发布时间】：2014-03-31 07:05:03
【问题描述】：

我想制作一个应用程序来查找两个字符串之间的差异。我该如何解决？

st  = "this is a cat.this is my cat."

st1 = "this is cat. this my cat."

输出应该是“a is”作为缺失的单词。

这是我的代码

@SuppressLint("DefaultLocale")
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    String st="this is a cat. this is my cat.";
    TextView tv1=(TextView) findViewById(R.id.textView2);

    String st1="This is cat. this my cat.";

    String blank="";
    if(st.toLowerCase().contains(st1.toLowerCase()))
    {
        st=st.toLowerCase().replace(st1.toLowerCase(), blank);
        tv1.setText(st);
    }
}

【问题讨论】：

但是“这是一只猫。这是我的猫。”不包含“这是猫。这是我的猫。”
我明白但我不知道我用什么方法来解决这个问题
好的，在下面查看我的答案
好的，非常感谢
你已经有 3 个答案了……你知道怎么做吗？

标签： android string string-matching

【解决方案1】：

你需要一种 DIFF 函数。

检查以下答案： Extract the difference between two strings in Java

How to perform string Diffs in Java?

【讨论】：

【解决方案2】：

你可以使用StringUtils.differences，它的来源：

6020        public static String difference(String str1, String str2) {
6021            if (str1 == null) {
6022                return str2;
6023            }
6024            if (str2 == null) {
6025                return str1;
6026            }
6027            int at = indexOfDifference(str1, str2);
6028            if (at == INDEX_NOT_FOUND) {
6029                return EMPTY;
6030            }
6031            return str2.substring(at);
6032        }
6033    
6034        /**
6035         * <p>Compares two Strings, and returns the index at which the
6036         * Strings begin to differ.</p>
6037         *
6038         * <p>For example,
6039         * <code>indexOfDifference("i am a machine", "i am a robot") -> 7</code></p>
6040         *
6041         * <pre>
6042         * StringUtils.indexOfDifference(null, null) = -1
6043         * StringUtils.indexOfDifference("", "") = -1
6044         * StringUtils.indexOfDifference("", "abc") = 0
6045         * StringUtils.indexOfDifference("abc", "") = 0
6046         * StringUtils.indexOfDifference("abc", "abc") = -1
6047         * StringUtils.indexOfDifference("ab", "abxyz") = 2
6048         * StringUtils.indexOfDifference("abcde", "abxyz") = 2
6049         * StringUtils.indexOfDifference("abcde", "xyz") = 0
6050         * </pre>
6051         *
6052         * @param str1  the first String, may be null
6053         * @param str2  the second String, may be null
6054         * @return the index where str2 and str1 begin to differ; -1 if they are equal
6055         * @since 2.0
6056         */
6057        public static int indexOfDifference(String str1, String str2) {
6058            if (str1 == str2) {
6059                return INDEX_NOT_FOUND;
6060            }
6061            if (str1 == null || str2 == null) {
6062                return 0;
6063            }
6064            int i;
6065            for (i = 0; i < str1.length() && i < str2.length(); ++i) {
6066                if (str1.charAt(i) != str2.charAt(i)) {
6067                    break;
6068                }
6069            }
6070            if (i < str2.length() || i < str1.length()) {
6071                return i;
6072            }
6073            return INDEX_NOT_FOUND;
6074        }
6075    
6076        /**
6077         * <p>Compares all Strings in an array and returns the index at which the
6078         * Strings begin to differ.</p>
6079         *
6080         * <p>For example,
6081         * <code>indexOfDifference(new String[] {"i am a machine", "i am a robot"}) -> 7</code></p>
6082         *
6083         * <pre>
6084         * StringUtils.indexOfDifference(null) = -1
6085         * StringUtils.indexOfDifference(new String[] {}) = -1
6086         * StringUtils.indexOfDifference(new String[] {"abc"}) = -1
6087         * StringUtils.indexOfDifference(new String[] {null, null}) = -1
6088         * StringUtils.indexOfDifference(new String[] {"", ""}) = -1
6089         * StringUtils.indexOfDifference(new String[] {"", null}) = 0
6090         * StringUtils.indexOfDifference(new String[] {"abc", null, null}) = 0
6091         * StringUtils.indexOfDifference(new String[] {null, null, "abc"}) = 0
6092         * StringUtils.indexOfDifference(new String[] {"", "abc"}) = 0
6093         * StringUtils.indexOfDifference(new String[] {"abc", ""}) = 0
6094         * StringUtils.indexOfDifference(new String[] {"abc", "abc"}) = -1
6095         * StringUtils.indexOfDifference(new String[] {"abc", "a"}) = 1
6096         * StringUtils.indexOfDifference(new String[] {"ab", "abxyz"}) = 2
6097         * StringUtils.indexOfDifference(new String[] {"abcde", "abxyz"}) = 2
6098         * StringUtils.indexOfDifference(new String[] {"abcde", "xyz"}) = 0
6099         * StringUtils.indexOfDifference(new String[] {"xyz", "abcde"}) = 0
6100         * StringUtils.indexOfDifference(new String[] {"i am a machine", "i am a robot"}) = 7
6101         * </pre>
6102         *
6103         * @param strs  array of strings, entries may be null
6104         * @return the index where the strings begin to differ; -1 if they are all equal
6105         * @since 2.4
6106         */
6107        public static int indexOfDifference(String[] strs) {
6108            if (strs == null || strs.length <= 1) {
6109                return INDEX_NOT_FOUND;
6110            }
6111            boolean anyStringNull = false;
6112            boolean allStringsNull = true;
6113            int arrayLen = strs.length;
6114            int shortestStrLen = Integer.MAX_VALUE;
6115            int longestStrLen = 0;
6116    
6117            // find the min and max string lengths; this avoids checking to make
6118            // sure we are not exceeding the length of the string each time through
6119            // the bottom loop.
6120            for (int i = 0; i < arrayLen; i++) {
6121                if (strs[i] == null) {
6122                    anyStringNull = true;
6123                    shortestStrLen = 0;
6124                } else {
6125                    allStringsNull = false;
6126                    shortestStrLen = Math.min(strs[i].length(), shortestStrLen);
6127                    longestStrLen = Math.max(strs[i].length(), longestStrLen);
6128                }
6129            }
6130    
6131            // handle lists containing all nulls or all empty strings
6132            if (allStringsNull || (longestStrLen == 0 && !anyStringNull)) {
6133                return INDEX_NOT_FOUND;
6134            }
6135    
6136            // handle lists containing some nulls or some empty strings
6137            if (shortestStrLen == 0) {
6138                return 0;
6139            }
6140    
6141            // find the position with the first difference across all strings
6142            int firstDiff = -1;
6143            for (int stringPos = 0; stringPos < shortestStrLen; stringPos++) {
6144                char comparisonChar = strs[0].charAt(stringPos);
6145                for (int arrayPos = 1; arrayPos < arrayLen; arrayPos++) {
6146                    if (strs[arrayPos].charAt(stringPos) != comparisonChar) {
6147                        firstDiff = stringPos;
6148                        break;
6149                    }
6150                }
6151                if (firstDiff != -1) {
6152                    break;
6153                }
6154            }
6155    
6156            if (firstDiff == -1 && shortestStrLen != longestStrLen) {
6157                // we compared all of the characters up to the length of the
6158                // shortest string and didn't find a match, but the string lengths
6159                // vary, so return the length of the shortest string.
6160                return shortestStrLen;
6161            }
6162            return firstDiff;
6163        }

然后

difference("this is a cat.this is my cat.", "this is cat. this my cat.");

如果你需要这个库中的更多方法，你可以只实现这个方法，或者整个库。

文档here。

【讨论】：

【解决方案3】：

简单的方法是在空格的基础上拆分两个字符串。例如

String[] separated_st = st.split(" ");
String[] separated_st1 = st1.split(" ");

现在你有两个数组。遍历它们，找出缺少的和没有的。

您也可以使用 StringTokenizer 类。希望这会有所帮助。

【讨论】：

【解决方案4】：

Try this one.

import java.awt.List;
import java.util.ArrayList;
import java.util.Arrays;


public class Classwithoutnewkeyword {
    public static void main(String args[]) {
        try {
                   String  s1  = "this is a cat.this is my cat.";
                   String  s2 = "this is cat. this my cat."
            String arr1[] = s1.split(" ");
            String arr2[] = s2.split(" ");
            java.util.List<String> list1 = new ArrayList<String>(
                    Arrays.asList(arr1));
            java.util.List<String> list2 = new ArrayList<String>(
                    Arrays.asList(arr2));

            ArrayList<String> tmp1 = new ArrayList<String>();
            ArrayList<String> tmp2 = new ArrayList<String>();

            for (int i = 0; i < arr1.length; i++) {
                int k = 0;
                for (int j = 0; j < arr2.length; j++) {

                    if (arr1[i].equalsIgnoreCase(arr2[j])) {
                        tmp1.add(arr1[i]);
                    } else {
                        tmp2.add(arr1[i]);
                    }

                }

            }
            for (String strs : tmp1) {
                list1.remove(strs);
            }

            System.out.print(list1.toString());

    }catch(Exception e)
    {
        e.printStackTrace();
    }
}
}

【讨论】：

在您的示例中，字符串 s1 的每个单词都与整个字符串 s2 进行比较，我想检查句子的顺序
所以你把字符串当作一个句子 ok va。然后应用 ma 逻辑，你肯定会得到你的解决方案。

【解决方案5】：

下面给出正确的解决方案：

import java.util.ArrayList;

public class Classwithoutnewkeyword {
    public static void main(String args[]) {
        try {
            String  s1  = "this is a cat. this is my cat.";
            String  s2 =  "this is cat. my cat.";
            String arr1[] = s1.split(" ");
            String arr2[] = s2.split(" ");
            ArrayList<String> tmp1 = new ArrayList<String>();
            ArrayList<String> tmp2 = new ArrayList<String>();

            for (int i = 0,j=0; i < arr1.length; i++) {
                if (arr1[i].equalsIgnoreCase(arr2[j])) {
                    tmp1.add(arr1[i]);
                    j++;
                } else {
                    tmp2.add(arr1[i]);
                }
            }
            System.err.println(tmp2);

        } catch(Exception e)
        {
            e.printStackTrace();
        }
    }
}

【讨论】：