[TwoPointers][2]Lintcode587. Two Sum - Unique pairs

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本文介绍了一种算法问题——寻找数组中和为目标值的唯一数对数量。通过两个指针法解决该问题，并考虑了重复元素的去重处理。提供了完整的Java实现代码。

Amazon | OA 2019 | Two Sum - Unique Pairs

Given an int array nums and an int target, find how many unique pairs in the array such that their sum is equal to target. Return the number of pairs.

Example 1:

Input: nums = [1, 1, 2, 45, 46, 46], target = 47
Output: 2
Explanation:
1 + 46 = 47
2 + 45 = 47

Example 2:

Input: nums = [1, 1], target = 2
Output: 1
Explanation:
1 + 1 = 2

Example 3:

Input: nums = [1, 5, 1, 5], target = 6
Output: 1
Explanation:
[1, 5] and [5, 1] are considered the same.

Solution1: TwoPointers

1.注意重复元素的去重
注意去重的方法；
如果用nums[start] == nums[start+1] start ++, 如果nums[start+1]还是一个重复的元素，nums[start+2]不是重复的元素，那么这种方法就不能去掉nums[start+1]。

public class Solution {
    /**
     * @param nums: an array of integer
     * @param target: An integer
     * @return: An integer
     */
    public int twoSum6(int[] nums, int target) {
        // write your code here
        
        if(nums == null || nums.length < 2) return 0;
        
        Arrays.sort(nums);
        
        int start = 0, end = nums.length -1;
        int sum = 0;
        
        while(start < end) {

            if(nums[start] + nums[end] == target){
                ++ sum;
                ++ start;
                -- end;
                while(start < end && nums[start] == nums[start - 1] ){
                    ++ start;
                }
                
                while(start < end && nums[end] == nums[end + 1] ){
                -- end;
                }
            }
            else if(nums[start] + nums[end] > target){
                -- end;
                
            }else {
                ++ start;
            }
            
        }
        
        return sum;
    }
}