How to solve the problem of exchanging digits and three steps in leetcode by golang

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Write a function that swaps the values of a and b in numbers = [a, b] without temporary variables.

Examples:

Input: numbers = [1,2]

Output: [2,1]

Tip:

numbers.length == 2

Solution:

Method 1:

summing

Method 2:

XOR

Code implementation:

func swapNumbers(numbers []int) []int { numbers[0]+=numbers[1] numbers[1]=numbers[0]-numbers[1] numbers[0]-=numbers[1] return numbers}func swapNumbers(numbers []int) []int { numbers[0]^=numbers[1] numbers[1]=numbers[0]^numbers[1] numbers[0]^=numbers[1] return numbers}

Three step problem. A child is climbing a staircase. The staircase has n steps. The child can climb one, two or three steps at a time. Implement a method to calculate how many ways a child can climb stairs. The result may be very large, you need to modulo 10000007 for the result.

Example 1:

Input: n = 3

Output: 4

Description: There are four ways to go

Example 2:

Input: n = 5

Output: 13

Tip:

n ranges between [1, 1000000]

Solution:

1, recursion

To reach step n, it can be 3 steps from step n-3, 2 steps from step n-2, or 1 step from step n-1.

2，dp

State transition equation, f (n)=f(n-3)+f(n-2)+f(n-1)

Since we're using n-3,n-2, n-1, the incremental approach

func waysToStep(n int) int { if n==1{ return 1 } if n==2{ return 2 } if n==3{ return 4 } return (waysToStep(n-3)+waysToStep(n-2)+waysToStep(n-1))00000007}func waysToStep(n int) int { dp:=make([]int,n) if n==1{ return 1 } if n==2{ return 2 } if n==3{ return 4 } dp[0]=1 dp[1]=2 dp[2]=4 for i:=3;i