字符串问题之 找到被指定的新类型字符

类似于 Ab  DC e

str="aaABCDEcBCg"   k=7  Ec     k=4 CD      k=10  g

解题思路:

  介绍一种最快最强的解题方法:

    从 k-1 位置开始 向左统计连续出现的大写字母数量 int  uNum    遇到小写字母就停止 

   如果 nNum为奇数, str[k..k+1]是被选中的新型字符

    如果nNum为偶数,且 str[k]是大写字母,str[k..k+1]是被选中的新型字符

    如果nNum为偶数,且srt[k]是小写字母,str[k]是被选中的新类型字符

package TT;



public class Test6 {
  
	public static String pointNewchar(String s, int k){
		if(s==null || s.equals("") || k<0 || k>=s.length()){
			return "";
		}
		
		char[] chas = s.toCharArray();
		int uNum = 0;
		for(int i = k-1; i>=0; i--){
			if( !isUpper(chas[i])){
				
				break;
					
			}
			
			uNum++;
			
		}
		
		if((uNum &1) ==1){    //判断是否是奇数
			 return s.substring(k-1, k+1);
			 
		}
		
		if(isUpper(chas[k])){
			return s.substring(k, k+2);
		}
		
		return String.valueOf(chas[k]);
		
		
	}
	
	
	public static boolean isUpper(char a){
		  
		if(a >'A' && a<'Z'){
			return true;
		}else{
			return false;
		}
	}
	
	
	
	public static void main(String[] args){
		 
		String str = "aaABCDEcBcg";
		int k = 7;
		String s = pointNewchar(str, k);
		System.out.println(s);
		
	}
	
	
	
	
}

  

 可以这么玩儿~

public class Test11 {

    public static int positionString(String str1, String str2) {
        String bigStr;
        String smallStr;
        int str1Length = str1.length();
        int str2Length = str2.length();
        if (str1Length >= str2Length) {
            bigStr = str1;
            smallStr = str2;
        } else {
            bigStr = str2;
            smallStr = str1;
        }

        int end = bigStr.length() - smallStr.length();
        for (int i = 0; i <= end; i++) {
            char now = bigStr.charAt(i);
            int smallIndex = 0;
            int bigIndex = i;
            if (now == smallStr.charAt(smallIndex)) {
                while ((smallIndex < (smallStr.length() - 1))
                        && bigStr.charAt(++bigIndex) == smallStr.charAt(++smallIndex))
                    ;
                if (smallIndex == (smallStr.length() - 1)) {
                    return i + 1;
                }
            }
        }
        return -1;

    }

    public static void main(String[] args) {
        String str1 = "aaABCDEcBCg";
        String str2 = "g";
        System.out.println(positionString(str1, str2));
    }

}
原文地址:https://www.cnblogs.com/toov5/p/7412125.html