Leetcode In JS #70 Climb Stairs

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Problem Description

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Analysis

分析

Solution

Solution 1 Recusive Version

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/**
 * @param {number} n
 * @return {number}
 */
/**
 * Dynamic Programming - Memorized Recursive Version
 */
var W = [0, 1, 2];

var climbStairs = function(n) {
    if (W[n] === undefined){
        W[n] = climbStairs(n - 2) + climbStairs(n - 1);
    }

    return W[n];
};

递归方式运行时间:

递归方式运行时间

Solution 2 Loop Version

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/**
 * @param {number} n
 * @return {number}
 */
/**
 * Dynamic Programming - Iteration Version
 */
var climbStairs = function(n) {
    var W = [0, 1, 2];
    for (var i = 3; i <= n; i++) {
        W[i] = W[i - 2] + W[i - 1];
    }

    return W[n];
};

循环方式运行时间:

循环方式运行时间

性能分析

理论上循环是比递归快的，因为同样是线性时间复杂度，递归调用需要频繁的进行栈操作，而循环不需要。