Input:
fn = function sum(a, b, c) { return a + b + c; }
inputs = [[1,2],[3]]
Output: 6
Explanation:
curriedSum(1, 2)(3) should return the same value as sum(1, 2, 3).
Input
"sum"
["call","call","getCallCount","call","getCallCount"]
[[2,2],[2,2],[],[1,2],[]]
Output
[4,4,1,3,2]
Explanation
const sum = (a, b) => a + b;
const memoizedSum = memoize(sum);
memoizedSum(2, 2); // Returns 4. sum() was called as (2, 2) was not seen before.
memoizedSum(2, 2); // Returns 4. However sum() was not called because the same inputs were seen before.
// Total call count: 1
memoizedSum(1, 2); // Returns 3. sum() was called as (1, 2) was not seen before.
// Total call count: 2
Input: fn = (a,b,c) => (a * b * c), calls = [[5,7,4],[2,3,6],[4,6,8]]
Output: [{"calls":1,"value":140}]
Explanation:
const onceFn = once(fn);
onceFn(5, 7, 4); // 140
onceFn(2, 3, 6); // undefined, fn was not called
onceFn(4, 6, 8); // undefined, fn was not called
Input: functions = [x => x + 1, x => x * x, x => 2 * x], x = 4
Output: 65
Explanation:
Evaluating from right to left …
Starting with x = 4.
2 * (4) = 8
(8) * (8) = 64
(64) + 1 = 65
Input:
nums = [1,2,3,4]
fn = function sum(accum, curr) { return accum + curr * curr; }
init = 100
Output: 130
Explanation:
initially, the value is init=100.
(100) + nums[0]^2 = 101
(101) + nums[1]^2 = 105
(105) + nums[2]^2 = 114
(114) + nums[3]^2 = 130
The final answer is 130.
Example 1:
Input: arr = [0,10,20,30], fn = function greaterThan10(n) { return n > 10; }
Output: [20,30]
Explanation:
const newArray = filter(arr, fn); // [20, 30]
The function filters out values that are not greater than 10
Example 2:
Input: arr = [1,2,3], fn = function firstIndex(n, i) { return i === 0; }
Output: [1]
Explanation:
fn can also accept the index of each element
In this case, the function removes elements not at index 0
Input: arr = [1,2,3], fn = function plusI(n, i) { return n + i; }
Output: [1,3,5]
Explanation: The function increases each value by the index it resides in.
Input: init = 0, calls = ["increment","increment","decrement","reset","reset"]
Output: [1,2,1,0,0]
Explanation:
const counter = createCounter(0);
counter.increment(); // 1
counter.increment(); // 2
counter.decrement(); // 1
counter.reset(); // 0
counter.reset(); // 0
Input:
n = -2
["call","call","call","call","call"]
Output: [-2,-1,0,1,2]
Explanation: counter() initially returns -2. Then increases after each sebsequent call.
Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"
Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.
Example 1:
Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32
Example 2:
Input: n = "82734"
Output: 8
82734-11111-11111-10111-10101-10100-10100-10100-10000
==>要做八次才能使得82734變成00000
Example 3:
Input: n = "27346209830709182346"
Output: 9
(換言之就是找到最大的數字)
const array1 = [1, 2, 3, 4];
// 0 + 1 + 2 + 3 + 4
const initialValue = 0;
const sumWithInitial = array1.reduce(
(accumulator, currentValue) => accumulator + currentValue,
initialValue
);
Input: nums = [10,4,8,3]
Output: [15,1,11,22]
Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].