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顺次数-1291 #116

@sl1673495

Description

@sl1673495

我们定义「顺次数」为:每一位上的数字都比前一位上的数字大 1 的整数。

请你返回由  [low, high]  范围内所有顺次数组成的 有序 列表(从小到大排序)。

示例 1:

输出:low = 100, high = 300
输出:[123,234]
示例 2:

输出:low = 1000, high = 13000
输出:[1234,2345,3456,4567,5678,6789,12345]

提示:

10 <= low <= high <= 10^9

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/sequential-digits
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

思路

先通过 lowhigh之间的长度对比,得到总共有几种数字长度。

然后循环这些数字长度 len 作为目标, 1 作为最小起点, 10 - len 作为最大的起点,不断尝试构造出长度为 len 的顺位数字即可。

/**
 * @param {number} low
 * @param {number} high
 * @return {number[]}
 */
let sequentialDigits = function (low, high) {
  let lowLen = low.toString().length
  let highLen = high.toString().length
  let lens = []
  for (let i = lowLen; i <= highLen; i++) {
    lens.push(i)
  }

  let res = []
  for (let i = 0; i < lens.length; i++) {
    let len = lens[i]
    for (let start = 1; start <= 10 - len; start++) {
      let num = start

      for (let n = start + 1; n < start + len; n++) {
        num = 10 * num + n
      }
      if (num <= high && num >= low) {
        res.push(num)
      }
    }
  }

  return res
}

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