three number sum

refer to: https://www.algoexpert.io/questions/Three%20Number%20Sum

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Three number Sum

• problem statement

给定一个数组和一个目标值，返回二维数组包含所有的三元组使得这三个数值的和等于目标值，返回的二维数组必须是数组内数组外升序排列。

• Solution: sort && pointers

首先给数组排序，然后遍历整个数组，固定当前索引的元素，然后设置左指针右指针

1. sort the array

2. traverse the entire array, fix one current value with index i, set left point(i+1) and right point(array.length) to position the another values. If array[i] + left + right = target, add the array into the result array。

• Code. (O(n^2), O(n))

notes：

1. 不要忘了while循环
2. 遍历数组的时候，i < array.length - 2.
3. 初始化二元数组结果集： 

   List<Integer[]> res = new ArrayList<Integer[]>();

def threeNumberSum(array, targetSum):
    array.sort()
    res = []
    for curr in range(len(array)-2):
        left = curr+1
        right = len(array)-1
        while left < right:
            currSum = array[curr] + array[left] + array[right]
            if currSum == targetSum:
                res.append([array[curr], array[left], array[right]])
                left += 1;
                right -= 1;
            elif currSum > targetSum:
                right -= 1
            elif currSum < targetSum:
                left += 1
    return res
        

class Program {
  public static List<Integer[]> threeNumberSum(int[] array, int targetSum) {
    Arrays.sort(array);
        List<Integer[]> res = new ArrayList<Integer[]>();
        for(int i = 0; i < array.length - 2; i++){
            int left = i+1, right = array.length - 1;
            while(left < right){
                if(array[i] + array[left] + array[right] == targetSum){
                    Integer[] triplet = {array[i], array[left], array[right]};
                    res.add(triplet);
                    left++;
                    right--;                
              }else if(array[i] + array[left] + array[right] < targetSum){
                    left++;
              }else if(array[i] + array[left] + array[right] > targetSum){
                  right--;
              }
            }
            
        }
    return res;
  }
}