LeetCode0-Two Sum-[Easy]
2263 ワード
postを恥ずかしがるが、勇敢に恥をかく.本シリーズはスラグの成長の道を記録し、目的は主に個人のブラシ問題の構想を総括することである.現在、language:Pythonがデフォルトです.
タイトルDescription:Two Sum
Given an array of integers, returnindicesof the two numbers such that they add up to a specific target. You may assume that each input would haveexactlyone solution, and you may not use thesameelement twice. Example: Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0,1].
考え方1:
遍歴加和.
結果:Time Limit Exceeded
考え方2:
考え方3の方法を参考にして,インデックスを巡る.
考え方3:(solutionを参照、ソースは図のように)
私はまだこの答えを理解していません......
タイトルDescription:Two Sum
Given an array of integers, returnindicesof the two numbers such that they add up to a specific target. You may assume that each input would haveexactlyone solution, and you may not use thesameelement twice. Example: Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0,1].
考え方1:
遍歴加和.
class Solution(object):
def twoSum(self, nums, target):
a = 0;
while (a < len(nums) ):
for x in nums:
for y in nums:
result = x + y
if (result == target ) & (nums.index(x) <> nums.index(y)):
return (nums.index(x),nums.index(y))
a = a + 1
結果:Time Limit Exceeded
考え方2:
考え方3の方法を参考にして,インデックスを巡る.
class Solution(object):
def twoSum(self, nums, target):
for x in range(len(nums)):
if target - nums[x] in nums:
if x <> nums.index(target - nums[x]):
return [x,nums.index(target - nums[x])]
考え方3:(solutionを参照、ソースは図のように)
私はまだこの答えを理解していません......
/**
* 。 , 。
* - , 。
* - : , , ,Java ,Android ,Big Data ,
* - :http://www.jiuzhang.com/?source=code
*/
public class Solution {
/*
* @param numbers : An array of Integer
* @param target : target = numbers[index1] + numbers[index2]
* @return : [index1 + 1, index2 + 1] (index1 < index2)
numbers=[2, 7, 11, 15], target=9
return [1, 2]
*/
public int[] twoSum(int[] numbers, int target) {
HashMap map = new HashMap<>();
for (int i = 0; i < numbers.length; i++) {
if (map.get(numbers[i]) != null) {
int[] result = {map.get(numbers[i]) + 1, i + 1};
return result;
}
map.put(target - numbers[i], i);
}
int[] result = {};
return result;
}
}