[leetcode] Two Sum

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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

package com.wyt.leetcodeOJ;import java.util.Arrays;import java.util.HashMap;import java.util.Map;/** * @author wangyitao * @version 1.0 *  */public class Two_sum {/** * @param args<br> * <p>测试例子<br> * Input: numbers={2, 7, 11, 15}, target=9<br> * Output: index1=1, index2=2 */public static void main(String[] args) {int[] numbers = {2, 3, 11, 15, 16, 17, 18, 19, 7};int target = 9;int[] a = twoSum(numbers, target);System.out.println(Arrays.toString(a));}/** * @param numbers:带求解int数组 * @param target:目标值 * @return 返回一个长度为2的int数组，两个index对应numbers元素之和为target的值 */private static int[] twoSum(int[] numbers, int target) {int len = numbers.length;    int[] index = new int[2];    /**     * 思路：对于numbers中的一个元素a，它的期待值是target-a     * 以target-a为键，a为值，将其放入map，每次put查找key是否存在，存在的话     * 说明存在这一的一个two sum     */Map<Integer, Integer> map = new HashMap<Integer, Integer>();for (int i = 0; i < len; i++) {if (!map.containsKey(numbers[i])) {//做一次查找，若存在就说明找到了数对map.put(target - numbers[i], i);//否则，放进map，继续找} else {index[0] = map.get(numbers[i]) + 1;index[1] = i + 1;}}return index;}}

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