Algorithm: Get all symmetric numbers (palindrome numbers) between 1-10000

topic

Get all symmetric numbers (palindrome numbers) between 1-n, for example:

1-9
11 22 33 ..
101 111 121 131 ...
....

Idea one

Traverse, remember two variables n1, n2.

  • n1 is the traversed number converted to a string
  • n2 is to convert each number to a string, then to an array, then reverse, and then to a string.

Comparing n1 and n2 for equality is a palindrome.

/**
 * 查询 1-max 的所有对称数(数组反转)
 * @param max 最大值
 */
export function findPalindromeNumbers1(max: number): number[] {
    
    
    const res: number[] = []
    if (max <= 0) return res

    for (let i = 1; i <= max; i++) {
    
    
        // 转换为字符串,转换为数组,再反转,比较
        const s = i.toString()
        if (s === s.split('').reverse().join('')) {
    
    
            res.push(i)
        }
    }

    return res
}

Idea two

Similar to idea 1, idea 1 is converted to array comparison, and idea 2 compares strings from head to tail

/**
 * 查询 1-max 的所有对称数(字符串前后比较)
 * @param max 最大值
 */
export function findPalindromeNumbers2(max: number): number[] {
    
    
    const res: number[] = []
    if (max <= 0) return res

    for (let i = 1; i <= max; i++) {
    
    
        const s = i.toString()
        const length = s.length

        // 字符串头尾比较
        let flag = true
        let startIndex = 0 // 字符串开始
        let endIndex = length - 1 // 字符串结束
        while (startIndex < endIndex) {
    
    
            if (s[startIndex] !== s[endIndex]) {
    
    
                flag = false
                break
            } else {
    
    
                // 继续比较
                startIndex++
                endIndex--
            }
        }

        if (flag) res.push(i)
    }

    return res
}

Idea three

generate rollovers

/**
 * 查询 1-max 的所有对称数(翻转数字)
 * @param max 最大值
 */
export function findPalindromeNumbers3(max: number): number[] {
    
    
    const res: number[] = []
    if (max <= 0) return res

    for (let i = 1; i <= max; i++) {
    
    
        let n = i
        let rev = 0 // 存储翻转数

        // 生成翻转数
        while (n > 0) {
    
    
            rev = rev * 10 + n % 10
            n = Math.floor(n / 10)
        }

        if (i === rev) res.push(i)
    }

    return res
}

test

// 功能测试
console.info(findPalindromeNumbers3(200))

// 性能测试
console.time('findPalindromeNumbers1')
findPalindromeNumbers1(100 * 10000)
console.timeEnd('findPalindromeNumbers1') // 408ms

console.time('findPalindromeNumbers2')
findPalindromeNumbers2(100 * 10000)
console.timeEnd('findPalindromeNumbers2') // 53ms

console.time('findPalindromeNumbers3')
findPalindromeNumbers3(100 * 10000)
console.timeEnd('findPalindromeNumbers3') // 42ms

  • Array conversion takes time
  • Manipulating numbers is faster, the prototype of the computer is the calculator

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Origin blog.csdn.net/weixin_43972437/article/details/130375934