标签：
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
1 public int[] twoSum(int[] nums, int target) {
2 if(nums.length == 0)
3 return null;
4
5 for (int i = 0; i < nums.length; i ++){
6 for (int j = i+1; j < nums.length; j ++){
7 if (nums[i] + nums[j] == target){
8 int[] result = new int[]{i+1, j+1};
9 return result;
10 }
11 }
12 }
13 return null;
14 }
1 public int[] twoSum(int[] nums, int target) {
2 if(nums.length == 0)
3 return null;
4
5 int[] result = new int[2];
6
7 int[] copyNums = new int[nums.length];
8 System.arraycopy(nums, 0, copyNums, 0, nums.length);
9 Arrays.sort(copyNums);
10
11 int low = 0;
12 int high = copyNums.length-1;
13
14 while (low < high){
15 int sum = copyNums[low] + copyNums[high];
16 if (sum == target){
17 result[0] = copyNums[low];
18 result[1] = copyNums[high];
19 break;
20 }
21 else if (sum < target)
22 low ++;
23 else
24 high --;
25 }
26
27 if (result.length == 0)
28 return null;
29
30 else{
31 int index1 = -1, index2 = -1;
32 for (int i = 0; i < nums.length; i ++){
33 if (result[0] == nums[i] && index1 == -1)
34 index1 = i + 1;
35 else if (result[1] == nums[i] && index2 == -1)
36 index2 = i + 1;
37 }
38
39 result[0] = index1;
40 result[1] = index2;
41 Arrays.sort(result);
42 return result;
43 }
44 }
1 public int[] twoSum(int[] nums, int target) {
2 if(nums.length < 2 || nums == null)
3 return null;
4
5 HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
6 int[] res = new int[2];
7
8 for (int i = 0; i < nums.length; i ++){
9 if (!map.containsKey(target-nums[i]))
10 map.put(nums[i], i + 1);
11 else{
12 res[0] = map.get(target-nums[i]);
13 res[1] = i + 1;
14 break;
15 }
16 }
17 return res;
18 }
Note:
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
I feel this is much more difficult than the before 2 sum one. We need three pointers here. And I found a trick, with which you did not write tree times list.add(element).
Arrays.asList(a, b, c)
1 public List<List<Integer>> threeSum(int[] nums) {
2 List<List<Integer>> triples = new ArrayList();
3 if (nums.length < 3)
4 return triples;
5
6 Arrays.sort(nums);
7 int i = 0, last = nums.length - 1;
8 while (i < last) {
9 int a = nums[i], j = i+1, k = last;
10 while (j < k) {
11 int b = nums[j], c = nums[k], sum = a+b+c;
12 if (sum == 0) triples.add(Arrays.asList(a, b, c));
13 if (sum <= 0) while (nums[j] == b && j < k) j++;
14 if (sum >= 0) while (nums[k] == c && j < k) k--;
15 }
16 while (nums[i] == a && i < last) i++;
17 }
18 return triples;
19 }
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).
1 public int threeSumClosest(int[] nums, int target) { 2 if (nums.length < 3) 3 return Integer.MAX_VALUE; 4 5 Arrays.sort(nums); 6 int result = nums[0] + nums[1] + nums[2]; 7 8 int i = 0; 9 int last = nums.length - 1; 10 while (i < last){ 11 int a = nums[i]; 12 int j = i + 1; 13 int k = last; 14 while (j < k){ 15 int b = nums[j]; 16 int c = nums[k]; 17 int sum = a + b + c; 18 int diff = Math.abs(target - sum); 19 20 if (diff < Math.abs(target - result)){ 21 result = sum; 22 } 23 24 if (diff == 0) 25 return sum; 26 else if (sum > target){ 27 while (nums[k] == c && j < k) 28 k --; 29 } 30 else if (sum < target){ 31 while (nums[j] == b && j < k) 32 j ++; 33 } 34 } 35 while (nums[i] == a && i < k) 36 i ++; 37 } 38 return result; 39 40 }
Note:
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)
For this problem, use 4 sum nests 3 sum nesting 2 sum and then get the result. Make 3 pointer move forward from the begining and one pointer move backward. One thing we need to remember the i < nums.length-2 at lease leave the length -1, length-2 for other pointer.
1 public List<List<Integer>> fourSum(int[] nums, int target) { 2 List<List<Integer>> res = new ArrayList<List<Integer>>(); 3 if (nums.length < 4) 4 return res; 5 6 Arrays.sort(nums); 7 8 int i = 0; 9 int last = nums.length - 1 ; 10 HashSet<List<Integer>> set = new HashSet<List<Integer>>(); 11 int n1 = nums[0], n2 = nums[1], n3 = nums[2], n4 = nums[last]; 12 13 while (i < last){ 14 int j = i + 1; 15 n1 = nums[i]; 16 while (j < last){ 17 int m = j + 1; 18 int k = last; 19 n2=nums[j]; 20 while (m < k){ 21 n3=nums[m]; n4=nums[k]; 22 int sum = n1 + n2 + n3 + n4; 23 if (sum == target){ 24 if (!set.contains(Arrays.asList(n1,n2,n3,n4))){ 25 res.add(Arrays.asList(n1,n2,n3,n4)); 26 set.add(Arrays.asList(n1,n2,n3,n4)); 27 } 28 } 29 if (sum <= target){ 30 while (n3 == nums[m] && m < k) 31 m ++; 32 } 33 if (sum >= target){ 34 while (n4 == nums[k] && m < k) 35 k --; 36 } 37 } 38 while (n2 == nums[j] && j < last) 39 j ++; 40 } 41 while (n1 == nums[i] && i < last) 42 i ++; 43 } 44 return res; 45 }
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原文地址：http://www.cnblogs.com/timoBlog/p/4665717.html
