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E

1522012438

tags: String

看StringA是不是包括所有 StringB的字符.

#### Basic Implementation

- 比较一下大小, null.

- 然后用int[]来count chars from A, count[x]++. 再对照chars in B, count[x]--

- 如果 count[c] < 0, 就 false.

- O(n)

```

/*

Compare two strings A and B, determine whether A contains all of the characters in B.

The characters in string A and B are all Upper Case letters.

Example

For A = "ABCD", B = "ABC", return true.

For A = "ABCD" B = "AABC", return false.

Tags Expand

Basic Implementation String LintCode Copyright

*/

/*

Thoughts:

Loop over A, B and ++/-- chars

count arr should have no negative results

*/

public class Solution {

public boolean compareStrings(String A, String B) {

if (A == null || B == null || A.length() < B.length()) {

return false;

}

int[] count = new int[26];

for (char c : A.toCharArray()) {

count[c - 'A']++;

}

for (char c : B.toCharArray()) {

count[c - 'A']--;

if (count[c - 'A'] < 0) {

return false;

}

}

return true;

}

}

/*

Previous notes

Thinking process:

Count the number of occurance for StringA.

Count the number of occurance for StringB.

Check if all of StringB's char# <= StringA's char# at each index.

*/

public class Solution {

public boolean compareStrings(String A, String B) {

if (A == null || B == null || A.length() < B.length()) {

return false;

}

int[] countA = new int[26];

int[] countB = new int[26];

for (int i = 0; i < A.length(); i++) {

countA[A.charAt(i) - 'A']++;

}

for (int i = 0; i < B.length(); i++) {

countB[B.charAt(i) - 'A']++;

if (countB[B.charAt(i) - 'A'] > countA[B.charAt(i) - 'A']) {

return false;

}

}

return true;

}

}

```