请教一个大家今天上午刚面试的一个题目，谢谢！

[qq_29448025]

[鱿鱼ing]

根据自己思路写了一个，仅供参考，欢迎补充。

public static void main(String[] args) {
		Map<Character, Integer> map = new HashMap<>();//创建一个map<字母，出现次数>
		Character first=null;//创建一个char对象代表第一个出现重复的
		String a= "hello world";
		char[] charArray = a.trim().toCharArray();//获取字符串里所有字符
		for (char c : charArray) { //循环字符数组
			Integer num = map.get(c);//从map里获取次数，如为null，带表第一次出现
			if (num==null) {
				num=1;
			}else{
				if (first==null) {//如果不为空且上面声明的first对象为空，则是第一个出现重复的字母，将次数+1
					first=c;
				}
				num++;
			}
			map.put(c, num);
		}
		for (char c : map.keySet()) {
			Integer num = map.get(c);
			if (num>1) {    //次数大于1代表重复，打印出来
				System.out.println("字母"+c+"重复出闲"+map.get(c)+"次！");
			}
		}
		System.out.println("第一个重复的字母是："+first);
	}

[minemine0418]

希望能帮到你

[soton_dolphin]

		char[] chars =  "hello world".toCharArray();
		Map<Character, Integer> mappingCount = new HashMap<>();
		for(char c : chars){
			mappingCount.computeIfPresent(Character.valueOf(c),	(k,v) -> v  +1);
			mappingCount.putIfAbsent(Character.valueOf(c), 1);
		}
		System.out.println(mappingCount);

[soton_dolphin]

		char[] chars =  "hello world".toCharArray();
		Map<Character, Integer> mappingCount = new HashMap<>();
		for(char c : chars){
			mappingCount.merge(Character.valueOf(c), 1, (oldV, newV) -> oldV + 1);
		}
		System.out.println(mappingCount);

[自由自在_Yu]

去掉空格是str = str.replace(" ",""); trim()只能去掉首尾空格

[soton_dolphin]

没看到还要报告第一个重复的字母

char[] chars =  "hello world".toCharArray();
		Map<Character, Integer> mappingCount = new HashMap<>();
		LinkedList<Character> repeatedChars = new LinkedList<>();
		for(char c : chars){
			mappingCount.computeIfPresent(Character.valueOf(c), (k, v) -> {repeatedChars.offer(c); return v + 1;});
			mappingCount.putIfAbsent(Character.valueOf(c), 1);
		}
		System.out.println(mappingCount);
		System.out.println("first repeated character: " + repeatedChars.poll());

[追赶阳光]

str.replaceAll(" ", ""); 去掉所有空格，包括首尾、中间 。

[御命丹心君奉天]

引用 4 楼 soton_dolphin 的回复:

		char[] chars =  "hello world".toCharArray();
		Map<Character, Integer> mappingCount = new HashMap<>();
		for(char c : chars){
			mappingCount.merge(Character.valueOf(c), 1, (oldV, newV) -> oldV + 1);
		}
		System.out.println(mappingCount);

好牛，大神啊

[yannsPeng]

引用 12 楼 dear_Alice_moon 的回复:

[quote=引用 3 楼 soton_dolphin 的回复:]

		char[] chars =  "hello world".toCharArray();
		Map<Character, Integer> mappingCount = new HashMap<>();
		for(char c : chars){
			mappingCount.computeIfPresent(Character.valueOf(c),	(k,v) -> v  +1);
			mappingCount.putIfAbsent(Character.valueOf(c), 1);
		}
		System.out.println(mappingCount);

好厉害，写的好简洁。 [/quote] java8 666.

[「已注销」]

public class Info2Action { public static void main(String[] args) { Map<String, Integer> map = new HashMap<String, Integer>(); int cishu = 0; String heeloWord = "heeelo word"; char[] s = heeloWord.trim().toCharArray(); for (int i = 0; i < s.length; i++) { for (int j = 0; j < s.length; j++) { if (i != j) { if (s[i] == s[j]) { i = j; cishu++; map.put(String.valueOf(s[i]), cishu); } } } cishu = 0; } Info2Action.test(map); } public static void test(Map<String, Integer> parameters) { Set set = parameters.entrySet(); Map.Entry[] entries = (Map.Entry[]) set.toArray(new Map.Entry[set.size()]); for (int i = 0; i < entries.length; i++) { System.out.println("重复字母:" + entries[i].getKey().toString()); System.out.println("重复:" + entries[i].getValue().toString()+"次"); } System.out.println("第一个重复的是"+entries[0].getKey().toString()); } }

[「已注销」]

上面有一个错了，，现在更正一下，i=j

[「已注销」]

public class Info2Action { public static void main(String[] args) { Map<String, Integer> map = new HashMap<String, Integer>(); int cishu = 0; String heeloWord = "oheelo word"; char[] s = heeloWord.trim().toCharArray(); for (int i = 0; i < s.length; i++) { for (int j = 0; j < s.length; j++) { if (i != j) { if (s[i] == s[j]) { cishu++; map.put(String.valueOf(s[i]), cishu); } } } cishu = 0; } Info2Action.test(map); } public static void test(Map<String, Integer> parameters) { Set set = parameters.entrySet(); Map.Entry[] entries = (Map.Entry[]) set.toArray(new Map.Entry[set.size()]); for (int i = 0; i < entries.length; i++) { System.out.println("重复字母:" + entries[i].getKey().toString()); System.out.println("重复:" + entries[i].getValue().toString()+"次"); } System.out.println("第一个重复的是"+entries[0].getKey().toString()); } }

[_明月]

引用 3 楼 soton_dolphin 的回复:

		char[] chars =  "hello world".toCharArray();
		Map<Character, Integer> mappingCount = new HashMap<>();
		for(char c : chars){
			mappingCount.computeIfPresent(Character.valueOf(c),	(k,v) -> v  +1);
			mappingCount.putIfAbsent(Character.valueOf(c), 1);
		}
		System.out.println(mappingCount);

好厉害，写的好简洁。

[75闪光雷]

看到这题我第一个想法是用hashmap

[_明月]

请问一下，楼主去面试时，这个楼主是面试的几年工作经验？ 3年么？

[ljheee]

这个 面试题目，之前我也碰到了

[Alfred8899]

进来学习学习

[加权不平权]

   public static void main(String[] args) {
        String input = "\"hello world\"";
        int[] result = new int[128];
        boolean firstTime = true;
        for (char c : input.toCharArray()) {
            result[c]++;
            if (firstTime && result[c] > 1) {
                System.out.printf("first repeat character:%c\n", c);
                firstTime = false;
            }
        }
        for (int i = 0; i < result.length; i++) {
            if (result[i] > 1) {
                System.out.printf("repeat character:%c,tims:%d.\n", (char) i, result[i]);
            }
        }
    }