Leetcode: 167. Two Sum II - Input array is sorted (Easy)

从一个已经排序的数组中查找出两个数，使它们的和为 0

Input: numbers={2, 7, 11, 15}, target=9 Output: index1=1, index2=2

  1. public int[] twoSum(int[] numbers, int target){
  2.     int i = 0, j = numbers.length - 1;
  3.     while(i < j){
  4.         int sum = numbers[i] + numbers[j];
  5.         if(sum == target) return new int[]{i + 1, j + 1};
  6.         else if(sum < target ) i++;
  7.         else j--;
  8.     }
  9.     return null;
  10. }

Leetcode: 345. Reverse Vowels of a String (Easy)

Given s = “leetcode”, return “leotcede”.

  1. private HashSet<Character> vowels = new HashSet<>(Arrays.asList('a','e','i','o','u','A','E','I','O','U'));
  3. public String reverseVowels(String s) {
  4.     if(s.length() == 0) return s;
  5.     int i = 0, j = s.length() - 1;
  6.     char[] result = new char[s.length()];
  7.     while(i <= j){
  8.         char ci = s.charAt(i);
  9.         char cj = s.charAt(j);
  10.         if(!vowels.contains(ci)){
  11.             result[i] = ci;
  12.             i++;
  13.         } else if(!vowels.contains(cj)){
  14.             result[j] = cj;
  15.             j--;
  16.         } else{
  17.             result[i] = cj;
  18.             result[j] = ci;
  19.             i++;
  20.             j--;
  21.         }
  22.     }
  23.     return new String(result);
  24. }

Leetcode: 633. Sum of Square Numbers (Easy)

Input: 5 Output: True Explanation: 1 1 + 2 2 = 5

public boolean judgeSquareSum(int c) {
    int left = 0, right = (int) Math.sqrt(c);
    while(left <= right){
        int powSum = left * left + right * right;
        if(powSum == c) return true;
        else if(powSum > c) right--;
        else left++;
    }
    return false;
}

Leetcode: 680. Valid Palindrome II (Easy)

Input: “abca” Output: True Explanation: You could delete the character ‘c’.

public boolean validPalindrome(String s) {
    int i = 0, j = s.length() - 1;
    while(i < j){
        if(s.charAt(i) != s.charAt(j)){
            return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
        }
        i++;
        j--;
    }
    return true;
}

private boolean isPalindrome(String s, int l, int r){
    while(l < r){
        if(s.charAt(l) != s.charAt(r))
            return false;
        l++;
        r--;
    }
    return true;
}

补充 Leetcode: 125 Valid Palindrome I (Easy)
给定一个字符串，验证它是否是回文串，只考虑字母和数字字符，可以忽略字母的大小写

    public boolean isPalindrome(String s) {
        StringBuffer sgood = new StringBuffer();
        for(int i = 0; i < s.length(); i++){
            char ch = s.charAt(i);
            if(Character.isLetterOrDigit(ch)){
                sgood.append(Character.toLowerCase(ch));
            }
        }
        StringBuffer sgood_rev = new StringBuffer(sgood).reverse();
        return sgood.toString().equals(sgood_rev.toString());
    }