英文:
Replace character in a String with HashMap
问题
我有一些想法来做这个我在if语句部分卡住了，怎么样才能使比较起作用呢？这个示例的预期输出是：字符串"ones"用于测试映射"numberthree"。String s = "The string 1s is use 2 test the map number3.";HashMap<Integer, String> map = new HashMap<>();map.put(1, "one");map.put(2, "two");map.put(3, "three");for (int i = 0; i < s.length(); i++) {if (s.charAt(i) 等于 map.contains(1)) {s[i] = map.get(1);}}
英文:
I have some idea to do this
I'm stuck on the if statement part, how can I make the comparison work?
The expected output for this example is: The string ones is use two test the map numberthree.
String s = "The string 1s is use 2 test the map number3.";HashMap<Integer,String> map = new HashMap<>();map.put(1,"one");map.put(2,"two");map.put(3,"three");for (int i = 0; i < s.length(); i++){if (s.charAt(i) is equals to map.contains(1){s[i] = map.get(1);}}
答案1
得分: 2
你需要将字符转换为数值以检查是否为 HashMap 中的 key。更好的方法是使用 StringBuilder，因为在找到它在映射中时，你正在追加 String：
String s = "The string 1s is use 2 test the map number3.";HashMap<Integer, String> map = new HashMap<>();map.put(1, "one");map.put(2, "two");map.put(3, "three");StringBuilder sb = new StringBuilder();for (char c : s.toCharArray()) {if (Character.isDigit(c) && map.containsKey(Character.getNumericValue(c))) {sb.append(map.get(Character.getNumericValue(c)));} else {sb.append(c);}}System.out.println(sb.toString());
另一种解决方案是将字符存储为 key：
String s = "The string 1s is use 2 test the map number3.";HashMap<Character, String> map = new HashMap<>();map.put('1', "one");map.put('2', "two");map.put('3', "three");StringBuilder sb = new StringBuilder();for (char c : s.toCharArray()) {if (map.containsKey(c)) {sb.append(map.get(c));} else {sb.append(c);}}System.out.println(sb.toString());
输出：
The string ones is use two test the map numberthree.
英文:
You need to convert the character to the numeric value to check if it's a key in the HashMap. A better approach would be to use a StringBuilder since you are appending String in case it's found in the map:
String s = "The string 1s is use 2 test the map number3.";HashMap<Integer,String> map = new HashMap<>();map.put(1,"one");map.put(2,"two");map.put(3,"three");StringBuilder sb = new StringBuilder();for (char c : s.toCharArray()){if (Character.isDigit(c) && map.containsKey(Character.getNumericValue(c))){sb.append(map.get(Character.getNumericValue(c)));}else {sb.append(c);}}System.out.println(sb.toString());
Another solution would be to store characters as the key:
String s = "The string 1s is use 2 test the map number3.";HashMap<Character,String> map = new HashMap<>();map.put('1',"one");map.put('2',"two");map.put('3',"three");StringBuilder sb = new StringBuilder();for (char c : s.toCharArray()){if (map.containsKey(c)){sb.append(map.get(c));}else {sb.append(c);}}System.out.println(sb.toString());
Output:
The string ones is use two test the map numberthree.
答案2
得分: 1
你可以迭代映射并将映射键用其值替换为字符串。
for (Map.Entry<Integer, String> e : map.entrySet()) {s = s.replace(e.getKey().toString(), e.getValue());}
我建议你不要按字符在字符串中进行迭代，因为如果你想要替换像22这样的两位或更多位数，你会遇到问题。
示例：
String s = "The string 1s is use 2 test the map number3.";HashMap<Integer, String> map = new HashMap<>();map.put(1, "one");map.put(2, "two");map.put(3, "three");for (Map.Entry<Integer, String> e : map.entrySet()) {s = s.replace(e.getKey().toString(), e.getValue());}System.out.println(s);
输出：
The string ones is use two test the map numberthree.
英文:
You can iterate the map and replace map key by its value in String.
for (Map.Entry<Integer,String> e: map.entrySet()) {s = s.replace(e.getKey().toString(), e.getValue());}
I recommend you not iterate by character in string since if you want to replace two or more digit numbers like 22 you faced problem.
Demo:
String s = "The string 1s is use 2 test the map number3.";HashMap<Integer,String> map = new HashMap<>();map.put(1,"one");map.put(2,"two");map.put(3,"three");for (Map.Entry<Integer,String> e: map.entrySet()) {s = s.replace(e.getKey().toString(), e.getValue());}System.out.println(s);
Output:
The string ones is use two test the map numberthree.
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