Double pointer
[977] The square of an ordered array
Category | Difficulty | Likes | Dislikes |
---|---|---|---|
algorithms | Easy (72.54%). | 257 | – |
Given an array of integers, nums, sorted in non-decrement order, return a new array of squares for each number, and ask for non-decrement order as well.
Individual solution
var sortedSquares = function (nums) {
for (let i in nums) {
nums[i] = nums[i] * nums[i];
}
nums.sort((a, b) = > a - b);
return nums;
};
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through | 116 ms | 45.8 MB | JavaScript |
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Analysis of the
The individual solution is the violent solution. Because arrays are sorted in non-decrementing order, you can also try the double-pointer method. Arrays are sorted in non-decrement order, and the maximum value of the squared elements must be at either end. Define the subscripts left and right for the left and right endpoints of the array, and compare the square values of the two endpoints. Insert larger values at the beginning of the new array arr until left=right.
The second solution
var sortedSquares = function (nums) {
let [left, right] = [0, nums.length - 1];
let arr = [];
while (left <= right) {
if (nums[left] * nums[left] < nums[right] * nums[right]) {
arr.unshift(nums[right] * nums[right]);
right--;
} else{ arr.unshift(nums[left] * nums[left]); left++; }}return arr;
};
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through | 160 ms | 44.7 MB | JavaScript |
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Analysis of the
The time complexity of the algorithm is reduced. Alternatively, you can use the absolute value function math.abs ().
const left = Math.abs(nums[i])
const right = Math.abs(nums[j])
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[189] Rotate the array
Category | Difficulty | Likes | Dislikes |
---|---|---|---|
algorithms | Medium (45.38%). | 1042 | – |
Given an array, move the elements of the array k ** positions to the right, where k ** is non-negative.
Individual solution
var rotate = function (nums, k) {
while (k > nums.length) {
k = k - nums.length;
}
let arr = nums.splice(nums.length - k, k);
arr.reverse();
for (let item of arr) {
nums.unshift(item);
}
return nums;
};
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Analysis of the
The individual solution is the violent solution. Consider inverting the array in the first place. For example, nums = [1,2,3,4,5,6,7] and k = 3. Nums = [7,6,5,4,3,2,1]. The array elements need to be moved to the front of the array for easy operation. In this case, the sequence of the k-bit array elements 7,6,5 and the remaining array elements 4,3,2,1 is opposite to the order of the target result, so the two “small arrays” can be reversed again. Tips: Invert an array by introducing a double-pointer method to swap each element: Define a subscript at the beginning and end of the array, and define an intermediate value temp to store the value of the last element for easy swapping. Write a separate inversion of the array function, call three times, more concise and convenient. If k is larger than the length of the array, it’s essentially moving the length of the remaining digits after the element has been moved by a number of lengths.
The second solution
var rotate = function (nums, k) {
while (k > nums.length) {
k = k - nums.length;
}
const reverse = (arr, start, end) = > {
while (start < end) {
let temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
return nums;
};
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through | 96 ms | 44.9 MB | JavaScript |
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