重新排列字符串

1. 我第一次做法

       public String restoreString(String s, int[] indices) {
           if (s == null || indices == null) {
               return null;
           }
   
           char[] result = new char[indices.length];
           for (int i = 0; i < indices.length; i++) {
               result[indices[i]] = s.charAt(i);
           }
   
           return String.valueOf(result);
       }
   

2. 其他解法

    public static String restoreString1(String s, int[] indices) {
           if (s == null || indices == null) {
               return null;
           }
   
           char[] chars = s.toCharArray();
           for (int i = 0; i < chars.length; i++) {
               if (i != indices[i]) {
                   int targetIndex = indices[i];
                   char tmp = chars[targetIndex];
                   chars[targetIndex] = chars[i];
                   chars[i] = tmp;
   
                   int tmp2 = indices[targetIndex];
                   indices[targetIndex] = indices[i];
                   indices[i] = tmp2;
   
                   // 由于替換完成后，并不能確定，再繼續(xù)執(zhí)行遍歷會(huì)將剩余的元素完全替換，則重置便利游標(biāo)
                   // 舉例： "aiohn" {3,1,4,2,0}
                   i=0;
               }
           }
   
           return String.valueOf(chars);
       }