LeetCode 1 - Two Sum
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原题如下:
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.
最直接的方法是两层循环遍历数组,对数组中所有元素一一相加并比对结果,时间复杂度是O(n^2)。代码如下:
1 public int[] twoSum1(final int[] numbers, int target) { 2 for (int i = 0; i < numbers.length; i++) { 3 for (int j = i+1; j < numbers.length; j++) { 4 int sum = numbers[i] + numbers[j]; 5 if(sum == target){ 6 return new int[]{i,j}; 7 } 8 } 9 } 10 return null; 11 }
优化之后时间复杂度可以降为O(nlogn),具体方法是先对数组进行排序(nlogn),然后分别使用两个指针从数组头和数组尾开始查找(n),利用数组有序的特性可以在O(n)的时间复杂度内完成检索。代码如下:
1 public int[] twoSum2(final int[] numbers, int target) { 2 List<Integer> indexes = new ArrayList<Integer>(); 3 for(int i = 0; i < numbers.length ; i++){ 4 indexes.add(i); 5 } 6 Collections.sort(indexes, new Comparator<Integer>(){ 7 @Override 8 public int compare(Integer o1, Integer o2) { 9 return numbers[o1]-numbers[o2]; 10 } 11 }); 12 int i=0,j=numbers.length-1; 13 while(i < j){ 14 int sum = numbers[indexes.get(i)] + numbers[indexes.get(j)]; 15 if(sum == target){ 16 break; 17 }else if(sum < target){ 18 i++; 19 }else{ 20 j--; 21 } 22 } 23 if(i < j){ 24 int i1 = indexes.get(i); 25 int i2 = indexes.get(j); 26 return i1< i2? new int[]{i1,i2}:new int[]{i2,i1}; 27 } 28 return null; 29 }
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