leetcode01. Two SumII

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本文介绍了一种高效算法，用于查找整数数组中所有和大于特定目标值的数对，并返回这些数对的数量。该算法采用排序加双指针策略实现，时间复杂度为O(n log n)，空间复杂度为O(n)。

Given an array of integers, find how many pairs in the array such that their sum is bigger than a specific target number. Please return the number of pairs.

解法

排序+双指针

/**
 * @Author RenXintao
 * @Date 9/6/17
 * @Time O(nlogn) @Space O(n)
 * @Desc Given an array of integers, find how many pairs in the array such that their sum is bigger than a specific
 * target number. Please return the number of pairs.
 */
public class TwoSumII_01 {
    public int twoSum(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int left = 0, right = nums.length - 1;
        int ans = 0;
        while (left < right) {
            if (nums[left] + nums[right] > target) {
                ans += right - left;
                right--;
            } else {
                left++;
            }
        }

        return ans;
    }

    public static void main(String[] args) {
        TwoSumII_01 twoSumII01 = new TwoSumII_01();
//        int[] nums = {2, 7, 11, 15};
//        int target = 24;
        int[] nums = {5, 4, 3, 7, 8};
        int target = 9;
        int ret = twoSumII01.twoSum(nums, target);
        System.out.print(ret);
    }
}