标签：val ber contain 输出 数值 == tar while solution
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].
解法1，sort后，左标依次移动，右标自适应左移动
与题意不符，题意要求输出角标，sort后角标变化
1 public class Solution { 2 public int[] twoSum(int[] nums, int target) { 3 Arrays.sort(nums); 4 int f = 0; 5 int l = nums.length; 6 int flag = l - 1; 7 int[] res = new int[2]; 8 for (int i = 0; i < nums.length; i++) { 9 10 while (nums[i] + nums[flag] > target) { 11 flag--; 12 } 13 if (nums[i] + nums[flag] == target) { 14 res[0] = i; 15 res[1] = flag; 16 return res; 17 } 18 } 19 return res; 20 } 21 }
解法2：hashmap，key存储数值，value存储value。
1 public class Solution { 2 public int[] twoSum(int[] nums, int target) { 3 int[] res = new int[2]; 4 Map<Integer, Integer> map = new HashMap<Integer, Integer>(); 5 for (int i = 0; i < nums.length; i++) { 6 if (map.containsKey(target - nums[i])) { 7 res[0] = map.get(target- nums[i]); 8 res[1] = i; //逻辑要清晰，res[1] = map.get(nums[i])......
9 return res; 10 } else { 11 map.put(nums[i], i); 12 } 13 } 14 return res; 15 } 16 }
标签：val ber contain 输出 数值 == tar while solution
原文地址：http://www.cnblogs.com/hafgyyb/p/7090228.html
