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【LeetCode】算法修炼 --- Two Sum
Question:
Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
Question Tags:
Array , Hash Table
New Words:
add up to：总计达
indices：index的复数
zero-based：从零开始的
Solution Ideas:
思路一：
两层遍历法：对于数组中的某一个数，对它及他以后的某个数求和，若和与target相等，则可确定这两值为所找的。
思路二：
HashMap--Value-key法：求a+b=target,也就是判断a和target-a是否都在这个数组中，
遍历判断map中是否有数组中的某个值target-a，如果没有，则把a的key以value作为key存到map中，
如果有，则所求的a,b得出来了。所求的索引值也就是a,b的索引值
两种方法都可以确保index1<index2.
Solution Java code:
import java.util.Arrays;import java.util.HashMap;import java.util.Map;public class TwoSum { public static void main(String[] args) { int[] numbers={2, 7, 11, 15}; int target = 9; int[] twoSum = twoSum(numbers,target); System.out.println("two sum indices are " + twoSum[0] + " and " + twoSum[1]); int[] twoSum2 = twoSum2(numbers,target); System.out.println("two sum indices are " + twoSum2[0] + " and " + twoSum2[1]); }
//思路1 public static int[] twoSum(int[] numbers, int target) { int i,j,sum; int[] indices = new int[2]; outfor:for (i=0;i<numbers.length;i++){ for (j=i+1;j>i && j<numbers.length;j++){ sum = numbers[i]+numbers[j]; if (sum == target){ indices[0]=i+1; indices[1]=j+1; break outfor; } } } return indices; }
//思路2 public static int[] twoSum2(int[] numbers, int target) { int[] indices = new int[2]; Map<Integer,Integer> map = new HashMap<Integer,Integer>(); for (int i=0;i<numbers.length;i++){ if (map.containsKey(target-numbers[i])){ indices[0]=map.get(target-numbers[i]); indices[1]=i+1; break; } map.put(numbers[i],i+1); } return indices; }}
【LeetCode】算法修炼 --- Two Sum