leetcode two sum

　two sum

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].

twosum

• 方案一使用哈希表

  class Solution {
  public int[] twoSum(int[] nums, int target) {
      Map<Integer, Integer> map = new HashMap();
      for(int i = 0; i<nums.length; i++){
          map.put(nums[i], i);
      }
      for (int i = 0; i < nums.length; i++) {
          int complement = target - nums[i];
          if (map.containsKey(complement) && map.get(complement) != i) {
              return new int[] { i, map.get(complement) };
          }
      }
      return null;
  }
  }

• 方案二快排

  class Solution {
  public int[] twoSum(int[] nums, int target) {
  
      // corner cases
      if (nums == null || nums.length <= 1) {
          return null;
      }
      int[] nums2 = Arrays.copyOf(nums, nums.length);
      Arrays.sort(nums);
      int left = 0;
      int right = nums.length - 1;
      int a = 0;
      int b = 0;
      while (left < right) {
          long sum = (long) nums[left] + (long) nums[right];
          if (sum == target) {
              a = nums[left];
              b = nums[right];
              break;
          } else if (sum < target) {
              left += 1;
          } else {
              right -= 1;
          }
      }
      // find index 1
      for(int i = 0; i < nums2.length; i++){
          if(nums2[i] == a) {
              left = i;
              break;
          }
      }
      // find index 2
      for(int i = nums2.length - 1; i >= 0; i--){
          if(nums2[i] == b) {
              right = i;
              break;
          }
      }
      return new int[]{Math.min(left, right),Math.max(left, right)};
  
  
  }
  }