567. Permutation in String

判断字符串s2的连续子串是否包含s1的排列。

双指针，保证count值不为正的情况下，判断是否存在一个区间，使得长度恰好为s1

• 时间复杂度O(m + n)，空间复杂度O(1)
• Runtime: 84 ms, faster than 95.47%
• Memory Usage: 40.2 MB, less than 92.33%

/**
 * @param {string} s1
 * @param {string} s2
 * @return {boolean}
 */
var checkInclusion = function(s1, s2) {
  const m = s1.length;
  const n = s2.length;
  if (m > n) {
    return false;
  }
  const count = new Array(26).fill(0);
  for (let i = 0; i < m; i++) {
    count[s1[i].charCodeAt() - 'a'.charCodeAt()]--;
  }
  let left = 0;
  for (let right = 0; right < n; right++) {
    const cur = s2[right].charCodeAt() - 'a'.charCodeAt();
    count[cur]++;
    
    while(count[cur] > 0) {
      count[s2[left].charCodeAt() - 'a'.charCodeAt()]--;
      left++;
    }
    if (right - left + 1 === m) {
      return true;
    }
  }
  return false;
};