2022-05-17

283. Move Zeroes

Description

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2:

Input: nums = [0]
Output: [0]

Constraints:

• 1 <= nums.length <= 104

• -231 <= nums[i] <= 231 - 1

Follow up: Could you minimize the total number of operations done?

Solution

func moveZeroes(nums []int) {
    for i, j := 0, 0; j < len(nums); {
        if nums[j] != 0 {
            nums[i], nums[j] = nums[j], nums[i]
            i++
        }
        j++
    }
}

167. Two Sum II - Input Array Is Sorted

Description

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.

Example 1:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

Constraints:

• 2 <= numbers.length <= 3 * 104

• -1000 <= numbers[i] <= 1000

• numbers is sorted in non-decreasing order.

• -1000 <= target <= 1000

• The tests are generated such that there is exactly one solution.

Solution

Approach #0

func twoSum(numbers []int, target int) []int {
    for i := 0; i < len(numbers); i++ {
        for j := i + 1; j < len(numbers); j++ {
            if numbers[i]+numbers[j] == target {
                return []int{i + 1, j + 1}
            }
        }
    }
    return []int{}
}

Approach #1

func twoSum(numbers []int, target int) []int {
    for i, j := 0, len(numbers)-1; i < j; {
        sum := numbers[i] + numbers[j]
        if sum == target {
            return []int{i + 1, j + 1}
        }
        if sum > target {
            j--
        } else {
            i++
        }

    }
    return []int{}
}