public class Solution {
    /*
     * @param : the 1st string
     * @param : the 2nd string
     * @return: uncommon characters of given strings
     */
    public String concatenetedString(String s1, String s2) {
        if(s1 == null){
            return s2;
        }
        if(s2 == null){
            return s1;
        }

        int[] map = new int[126];
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < s2.length(); i++){
            map[s2.charAt(i)] = 1;
        }

        for(int i = 0; i < s1.length(); i++){
            if(map[s1.charAt(i)] == 0){
                sb.append(s1.charAt(i));
            }else{
                map[s1.charAt(i)] = 2;
            }
        }

        for(int i = 0; i < s2.length(); i++){
            if(map[s2.charAt(i)] == 1){
                sb.append(s2.charAt(i));
            }
        }
        return sb.toString();
    }
};
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0
L同学
发布于 11/1/2017, 10:26:33 PM
Initialize result as empty string.
Push all characters of 2nd string in map with count as 1.
Traverse first string and append all those characters to result that are not present in map. Characters that are present in map, make count 2.
Traverse second string and append all those characters to result whose count is 1.

代码
评论 0
/**
* 本参考程序来自九章算法，由 @L同学 提供。版权所有，转发请注明出处。
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* - 更多详情请见官方网站：http://www.jiuzhang.com/?source=code
*/

public class Solution {
    /*
     * @param : the 1st string
     * @param : the 2nd string
     * @return: uncommon characters of given strings
     */
    public String concatenetedString(String s1, String s2) {
        // write your code here
        StringBuilder sb = new StringBuilder();
        Map<Character, Integer> map = new HashMap<>();

        for (int i = 0; i < s2.length(); i++) {
            if (!map.containsKey(s2.charAt(i))) {
                map.put(s2.charAt(i), 1);
            }
        }

        for (int i = 0; i < s1.length(); i++) {
            if (!map.containsKey(s1.charAt(i))) {
                sb.append(s1.charAt(i));
            } else {
                map.put(s1.charAt(i), 2);
            }
        }

        for (int i = 0; i < s2.length(); i++) {
            if (map.get(s2.charAt(i)) == 1) {
                sb.append(s2.charAt(i));
            }
        }

        return sb.toString();
    }
};