Leetcode.com/problems/tw…

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9 Output: [0,1] Output: Because nums[0] + nums[1] == 9, we return [0,1].Copy the code

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]
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Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]
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Constraints:

  • 2 <= nums.length <= 104

  • -109 <= nums[i] <= 109

  • -109 <= target <= 109

  • Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

Solution a:

Brute force, order N ^ 2.

class Solution {
    fun twoSum(nums: IntArray, target: Int): IntArray {
        val result = mutableListOf<Int>()

        for (i in nums.indices) {
            val x = nums[i]
            for (j in i + 1 until nums.size) {
                val y = nums[j]
                if (x + y == target) {
                    result.add(i)
                    result.add(j)
                }
            }
        }
        return result.toIntArray()
    }
}
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Method 2:

Use Map to save subscripts and corresponding values of subscripts, complexity O(N).

class Solution {
    fun twoSum(nums: IntArray, target: Int): IntArray {
        val result = mutableListOf<Int>()
        val map = mutableMapOf<Int, Int>()

        for (index in nums.indices) {
            map[index] = nums[index]
        }

        for (index in nums.indices) {
            val x = nums[index]
            val y = target - x
            if (map.containsValue(y) && map.values.lastIndexOf(y) > index) {
                result.add(index)
                result.add(map.values.lastIndexOf(y))
            }
        }
        return result.toIntArray()
    }
}
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