Question

01.01. Is Unique

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Description

Implement an algorithm to determine if a string has all unique characters. What if you cannot use additional data structures?

Example 1:

Input:  = "leetcode"

Output: false

Example 2:

Input: s = "abc"

Output: true

Note:

• 0 <= len(s) <= 100

Solutions

Solution 1: Bit Manipulation

Based on the examples, we can assume that the string only contains lowercase letters (which is confirmed by actual verification).

Therefore, we can use each bit of a $32$-bit integer mask to represent whether each character in the string has appeared.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.

class Solution:
  def isUnique(self, astr: str) -> bool:
    mask = 0
    for c in astr:
      i = ord(c) - ord('a')
      if (mask >> i) & 1:
        return False
      mask |= 1 << i
    return True

class Solution {
  public boolean isUnique(String astr) {
    int mask = 0;
    for (char c : astr.toCharArray()) {
      int i = c - 'a';
      if (((mask >> i) & 1) == 1) {
        return false;
      }
      mask |= 1 << i;
    }
    return true;
  }
}

class Solution {
public:
  bool isUnique(string astr) {
    int mask = 0;
    for (char c : astr) {
      int i = c - 'a';
      if (mask >> i & 1) {
        return false;
      }
      mask |= 1 << i;
    }
    return true;
  }
};

func isUnique(astr string) bool {
  mask := 0
  for _, c := range astr {
    i := c - 'a'
    if mask>>i&1 == 1 {
      return false
    }
    mask |= 1 << i
  }
  return true
}

function isUnique(astr: string): boolean {
  let mask = 0;
  for (let j = 0; j < astr.length; ++j) {
    const i = astr.charCodeAt(j) - 'a'.charCodeAt(0);
    if ((mask >> i) & 1) {
      return false;
    }
    mask |= 1 << i;
  }
  return true;
}

/**
 * @param {string} astr
 * @return {boolean}
 */
var isUnique = function (astr) {
  let mask = 0;
  for (const c of astr) {
    const i = c.charCodeAt() - 'a'.charCodeAt();
    if ((mask >> i) & 1) {
      return false;
    }
    mask |= 1 << i;
  }
  return true;
};