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ClimbingStairs.cpp
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/*
Source: https://leetcode.com/problems/climbing-stairs/
Approach 1
Time: O(n), where n is the given n
Space: O(n), an array is needed to store the steps
*/
class Solution {
public:
int climbStairs(int n) {
int ways[n + 1];
ways[0] = 1;
ways[1] = 1;
for(int i = 2; i <= n; ++i) {
ways[i] = ways[i - 1] + ways[i - 2];
}
return ways[n];
}
};
-------------------------------------------------------------------------------------------------------
/*
Approach 2 (without using extra array)
Time: O(n), where n is the given n
Space: O(1), in-place
*/
class Solution {
public:
int climbStairs(int n) {
int first = 1;
int second = 1;
int ways = 0;
for(int i = 1; i < n; ++i) {
ways = first + second;
first = second;
second = ways;
}
return second;
}
};
-------------------------------------------------------------------------------------------------------
/*
Approach 3 (performing less operations (i.e. no swapping))
Time: O(n), where n is the given n
Space: O(1), in-place
*/
class Solution {
public:
int climbStairs(int n) {
int ways[] = {1, 1};
for(int i = 2; i <= n; ++i) {
ways[i % 2] = ways[0] + ways[1];
}
return ways[n % 2];
}
};
-------------------------------------------------------------------------------------------------------
/*
Approach 4 (same as Approach 3 but using bitwise operator)
Time: O(n), where n is the given n
Space: O(1), in-place
*/
class Solution {
public:
int climbStairs(int n) {
int ways[] = {1, 1};
for(int i = 2; i <= n; ++i) {
ways[i & 1] = ways[0] + ways[1];
}
return ways[n & 1];
}
};
----------------------------------------------------------------------------------------------------
/*
Approach 5 (making using of n only in loop)
Time: O(n), where n is the given n
Space: O(1), in-place
*/
class Solution {
public:
int climbStairs(int n) {
int first = 1;
int second = 1;
int steps = 0;
while(--n > 0) {
steps = first + second;
first = second;
second = steps;
}
return second;
}
};