LeetCode 01/10/18

起晚了起晚了，又玩了会儿炉石，堕落啊堕落。。
今天得把两个String的lec看完，把里面的题做完。

347. Top K Frequent Elements
     这题思路很简单 就是先存hashMap 再用heap来sort求top frequent
     但是implementation十分蛋疼，首先是要把map里的pair存入heap
     for(Map.Entry entry : map.entrySet())
     在heap中比较时，还得用到lambda
     PriorityQueue> pq = new PriorityQueue<>((a, b) -> a.getValue()-b.getValue());
     反正很蛋疼，有缘学下lambda和再写一遍吧。。✘✘！

348. Missing Number
     way1 hashset way2 sum way3 binary search way4 XOR

Common Numbers Of Two Sorted Arrays
way1 hashset O(m + n) space O(min(m, n))
way2 2pointer 谁小移谁 O(m + n) space O(1)
way3 binary search 如果两个array 大小相差悬殊 O(mlogn)

type 1 char removal

---

Remove Certain Characters
两个pointer来做 （not include）
remove需要inplace 不用交换直接覆盖即可
最后生成新的string可以用String(arr, 0, i)来完成
String(char[] value, int offset, int count)
Allocates a new String that contains characters from a subarray of the character array argument.

Remove Spaces
两个pointer来做
先将leading 和 duplicate去掉
分情况讨论： case1： if arr[j] != ' ', arr[i] = arr[j], i++,j++
case2 : if arr[j] == ' ',
case2.1: if j == 0, ignore
case2.2: if arr[j - 1] != ' ', retain
case2.3: if arr[j - 1] == ' ', ignore
做完上述操作后再加上处理最后trailing space的一步

 for (int j = 0; j < arr.length; j++) {
      // ignore space 2.1 2.3
      if (arr[j] == ' ' && (j == 0 || arr[j - 1] == ' ')) {
        continue;
      }
      // case1, 2.2
      arr[i] = arr[j];
      i++;
    }
    // post- processing: if have trailing sapce
    if (i > 0 && arr[i - 1] == ' ') {
      return new String(arr, 0, i - 1);
    }
return new String(arr, 0, i);

Type2 char de-duplication

---

Remove Adjacent Repeated Characters I
两个pointer

Remove Adjacent Repeated Characters IV
用一个stack来记录并比较是否重复
但其实可以直接用两个指针来做

// arr[i] as the top of a stack
    for (int j = 1; j < arr.length; j++) {
      if (i == -1 || arr[j] != arr[i]) {
        arr[++i] = arr[j];
      } else {
        // pop and ignore all the consecutive duplicates
        i--;
        while (j + 1 < arr.length && arr[j] == arr[j + 1]) {
          j++;
        }

Type3 sub str

---

Determine If One String Is Another's Substring

int n = large.length();
    int m = small.length();
    for (int i = 0; i <= n - m; i++) {
      int j = 0;
      while (j < m) {
        if (large.charAt(i + j) != small.charAt(j)) {
          break;
        }
        j++;
      }
      if (j == m) {
        return i;
      }
    }
    return -1;