1. 两数之和

两数之和 #

题目 #

给定一个整数数组 nums  和一个目标值 target,请你在该数组中找出和为目标值的那两个整数,并返回他们的数组下标。

你可以假设每种输入只会对应一个答案。但是,你不能重复利用这个数组中同样的元素。

示例:

给定 nums = [2, 7, 11, 15], target = 9

因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

分析 #

使用 hash 表,记录过程数据,空间换时间

题解 #




func twoSum(nums []int, target int) []int {
    tmp := make(map[int]int)
    for k, v := range(nums) {
        first, ok := tmp[v]
        if ok {
            return []int{first, k}
        }

        tmp[target - v] = k
    }

    return []int{}
}
class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap <Integer, Integer> tmp = new HashMap<Integer, Integer>();
        for(int i = 0; i < nums.length; i++) {
            if (tmp.get(nums[i]) != null) {
                return new int[] {tmp.get(nums[i]), i};
            } else {
                tmp.put(target - nums[i], i);
            }
        }

        return new int[] {};
    }
}


class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        r = {}
        for i in range(len(nums)):
            if nums[i] in r:
                return i, r[nums[i]]
            else:
                r[target - nums[i]] = i
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:






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