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.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.
【问题分析】
kSUM系列的问题有好多个,如下:
我们对这几个题目分别分析并进行总结。
【思路】
1. Two Sum
解决这个问题可以直接利用两层循环对数组进行遍历,这样的时间复杂度为O(N2)。一个巧妙的办法是利用java中的HashMap来解决这个问题,代码如下:
1 public class Solution { 2 public int[] twoSum(int[] nums, int target) { 3 int[] result = 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 result[1] = i; 8 result[0] = map.get(target - nums[i]); 9 return result; 10 } 11 map.put(nums[i], i); 12 } 13 return result; 14 } 15 }
由于HashMap的查询效率很高,HashMap的一些操作技巧:http://jiangzhenghua.iteye.com/blog/1196391
2. Two Sum II - Input array is sorted
这个two sum问题中,数组中的元素是已经排序的,我们从数组的头和尾向数组中间靠拢,如果头尾元素相加大于target,则尾指针向前移动一步,如果小于target,则头指针向后移动一步,直到两指针相遇或者相加结果为target。示例如下:[2,3,4,5] target = 7
思路很简单,代码如下:
1 public class Solution { 2 public int[] twoSum(int[] numbers, int target) { 3 int[] result = new int[2]; 4 if(numbers == null || numbers.length < 2) return result; 5 6 int left = 0, right = numbers.length-1; 7 while(left < right){ 8 int cur = numbers[left] + numbers[right]; 9 if(cur == target){ 10 result[0] = left+1; 11 result[1] = right+1; 12 return result; 13 } 14 else if(cur < target){ 15 left++; 16 } 17 else{ 18 right--; 19 } 20 } 21 return result; 22 } 23 }
可见,排序后的two sum效率还是很高的。
3. Three sum
The idea is to sort an input array and then run through all indices of a possible first element of a triplet. For each possible first element we make a standard bi-directional 2Sum sweep of the remaining part of the array. Also we want to skip equal elements to avoid duplicates in the answer without making a set or smth like that.
结合two sum和Two Sum II - Input array is sorted我们可以比较好解决这个问题。上面这段话的思路是:先对数组进行排序,然后遍历排序后数组,把每一个元素当做三元组的开始元素,剩下的两个元素的查找和Two Sum II是相同的。在这个过程中需要注意的就是去重,一些重复出现的元素要跳过。代码如下:
1 public class Solution { 2 public List<List<Integer>> threeSum(int[] nums) { 3 Arrays.sort(nums); 4 List<List<Integer>> result = new LinkedList<>(); 5 for(int i = 0; i < nums.length-2; i++){ 6 if(i == 0 || (i>0 && nums[i] != nums[i-1])){ 7 int target = 0 - nums[i]; 8 int left = i + 1; 9 int right = nums.length - 1; 10 while(left < right){ 11 if(nums[left] + nums[right] == target){ 12 result.add(Arrays.asList(nums[i], nums[left], nums[right])); 13 while(left < right && nums[left] == nums[left+1]) left++; 14 while(left < right && nums[right] == nums[right-1]) right--; 15 left++; right--; 16 } 17 else if (nums[left] + nums[right] < target) 18 left ++; 19 else right--; 20 } 21 } 22 } 23 return result; 24 } 25 }
4. 3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
这个题目在3sum的基础上做了一点变化,要在所有2元组中找到与目标值最接近的三元组的和。我的思路和上一个题目类似,先对数组进行排序,然后在遍历过程中如果发现了更接近的元组的和,则更新最接近的值。如果发现了和值有和目标值相等的,则直接返回目标值。代码如下:
1 public class Solution { 2 public int threeSumClosest(int[] nums, int target) { 3 Arrays.sort(nums); 4 int closest = nums[0]+nums[1]+nums[2]; 5 for(int i = 0; i < nums.length-2; i++){ 6 int left = i+1, right = nums.length-1; 7 while(left < right){ 8 int cur = nums[i] + nums[left] + nums[right]; 9 if(cur == target) return cur; 10 else if(cur > target) right--; 11 else left++; 12 if(Math.abs(cur-target) < Math.abs(closest-target)) 13 closest = cur; 14 } 15 } 16 return closest; 17 } 18 }
5. 4sum
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note: The solution set must not contain duplicate quadruplets.
For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0. A solution set is: [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]
这个问题的解决可以借鉴3 sum的思路,只要在3sum外层再增加一层循环即可,代码如下:
1 public class Solution { 2 public List<List<Integer>> fourSum(int[] nums, int target) { 3 Arrays.sort(nums); 4 List<List<Integer>> result = new LinkedList<>(); 5 for(int i = 0; i < nums.length-3; i++){ 6 if(i == 0 || (i>0 && nums[i] != nums[i-1])){ 7 int curtarget1 = target - nums[i]; 8 for(int j = i+1; j < nums.length-2; j++){ 9 if(j == i+1 || (j>i+1 && nums[j] != nums[j-1])){ 10 int curtarget2 = curtarget1 - nums[j]; 11 int left = j + 1; 12 int right = nums.length - 1; 13 while(left < right){ 14 if(nums[left] + nums[right] == curtarget2){ 15 result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right])); 16 while(left < right && nums[left] == nums[left+1]) left++; 17 while(left < right && nums[right] == nums[right-1]) right--; 18 left++; right--; 19 } 20 else if (nums[left] + nums[right] < curtarget2) 21 left ++; 22 else right--; 23 } 24 } 25 } 26 } 27 } 28 return result; 29 } 30 }
自此,这几个解锁的N sum的题目就做完了,这种题目用回溯法适合不适合呢?
另外需要注意在求解的时候要去掉重复的解,如果排序后的元素是a,b,c,d,求解过程如果选定的元素和上一个选定的元素是相同的,则可以直接跳过该元素。至于为什么是这样,大家可以思考一下。