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| package cn.edu.xidian.sselab.array; import java.util.HashMap; import java.util.Map; /** * * @author zhiyong wang * title:Two Sum * content: * 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 * */ public class TwoSum { //最开始的，也是最简单的思路，就是遍历整个数组，时间复杂度O(n^2); //这里自己改进的地方，一个是第一层循环只用遍历到倒数第二个数，第二层循环从第二个数开始遍历；第二个是如果i>j，直接跳过这次比较 public int[] twoSum(int[] nums, int target){ int length = nums.length; int[] indices = new int[2]; for(int i=0;i<length-1;i++){ for(int j=1;j<length;j++){ if(i >= j){ continue; } if(nums[i]+nums[j]==target && i<j){ indices[0] = i + 1; indices[1] = j + 1; } } } return indices; } //参考比较好的，O(n)的时间复杂度 //从开始进行遍历，将遇到的数放入到一个Map中，记录下数值与他的下标， //用目标结果减去遍历的值，如果得到的结果在Map中存在，则说明存在，取出他们的下标返回即可 public int[] twoSums(int[] nums, int target){ int length = nums.length; int[] indices = new int[2]; Map<Integer,Integer> map = new HashMap<Integer,Integer>(); for(int i=0;i<length;i++){ if(map.containsKey(target - nums[i])){ indices[0] = map.get(target-nums[i]); indices[1] = i + 1; return indices; }else{ map.put(nums[i], i+1); } } return indices; } } ? |
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原文地址：http://www.cnblogs.com/wzyxidian/p/5074740.html
