查找字符串中重复字符,下标,重复个数..?

思路:一遍循环将字符串转换为字典,字典中保存想要的结果集
1,最外层是数组,保证输入顺序与字符串中出现顺序一致.
2,获取数组lastObject
3,比较lastObject的key,与当前字符是否相同, 相同:数据加
4,key不同,判断出现次数,lenth==1,替换, >1新增.

//1.声明一个共用数组,self.arrayM: NSMutableArray
//2.code
//3.kvc获取结果集
- (void)repeatInfoIn:(NSString *)string
{
        NSInteger count = string.length;
    if (count <= 1) return;
    for (NSInteger i = 0; i < string.length; i++) {
        NSString *ichar = [string substringWithRange:NSMakeRange(i, 1)];
        if (i == 0) {
            NSDictionary *valueDict = @{@"key": ichar, @"index": @(i), @"count": @(1), @"subStr": ichar};
            [self.arrayM addObject:valueDict];
            continue;
        }
        NSMutableDictionary *previousDict = [[self.arrayM lastObject] mutableCopy];
        NSString *previousKey = previousDict[@"key"];
        NSInteger previousCount = [previousDict[@"count"] integerValue];
        if ([previousKey isEqualToString:ichar]) {
            previousCount += 1;
            NSString *subStr = [NSString stringWithFormat:@"%@%@", previousDict[@"subStr"], ichar];
            previousDict[@"count"] = @(previousCount);
            previousDict[@"subStr"] = subStr;
            [self.arrayM replaceObjectAtIndex:(self.arrayM.count-1) withObject:previousDict];
        }else {
            if (previousCount == 1) {
                NSDictionary *valueDict = @{@"key": ichar, @"index": @(i), @"count": @(1), @"subStr": ichar};
                [self.arrayM replaceObjectAtIndex:(self.arrayM.count-1) withObject:valueDict];
            }else {
                NSDictionary *valueDict = @{@"key": ichar, @"index": @(i), @"count": @(1), @"subStr": ichar};
                [self.arrayM addObject:valueDict];
            }
        }
    }
}

- (void)touchesBegan:(NSSet<UITouch *> *)touches withEvent:(UIEvent *)event {
    [super touchesBegan:touches withEvent:event];
    NSString *string = @"asdqwezccdddd";
    [self repeatInfoIn:string];
    NSLog(@"%@", self.arrayM);
    NSArray *subStrArray = [self.arrayM valueForKeyPath:@"subStr"];
    NSLog(@"%@", subStrArray);
}

/* java中可能有更简便的方法,我这里只是一种思路推演. */
static void linkedHashMapOperate() {
        String str = "aaasdqwezccdddd";
        ArrayList<HashMap<Character,HashMap<String, Object>>> list = new ArrayList<>();
        for (int i = 0; i < str.length(); i++) {
            char key = str.charAt(i);
            int lastIndex = Math.max(0, list.size()-1);
            if (i == 0) {
                HashMap<Character, HashMap<String, Object>> hashMap = new HashMap<>();
                var value = new HashMap<String, Object>();
                value.put("index", i);
                value.put("lenth", 1);
                value.put("key", key);
                hashMap.put(key, value);
                list.add(hashMap);
                continue;
            }
            HashMap<Character, HashMap<String, Object>> hashMap = list.get(lastIndex);
            Set<Map.Entry<Character, HashMap<String, Object>>> entries = hashMap.entrySet();
            char lastKey = key;
            var value = new HashMap<String, Object>();
            for (Map.Entry<Character, HashMap<String, Object>> entry: entries) {
                lastKey = (char) entry.getKey();
                value = entry.getValue();
            }
            if (lastKey == key) {//说明上一个元素与当前元素重复
                int index = (int) value.get("index");
                int lenth = (int) value.get("lenth");
                index += 1;
                lenth += 1;
                value.put("index", index);
                value.put("lenth", lenth);
            }else {
                //判断上一个元素lenth==1
                int lenth = (int) value.get("lenth");
                if (lenth==1) {//替换--->key
                    value.put("key",key);
                    hashMap.remove(lastKey);
                    hashMap.put(key,value);
                    list.set(lastIndex, hashMap);
                }else {//新增
                    HashMap<Character, HashMap<String,Object>> nHashMap = new HashMap();
                    var nValue = new HashMap<String, Object>();
                    nValue.put("index", i);
                    nValue.put("lenth", 1);
                    nValue.put("key", key);
                    nHashMap.put(key, nValue);
                    list.add(nHashMap);
                }
            }
        }
        System.out.println(list);
    }