三数之和

题目：力扣

解题思路：

排序+双指针

排序使用的是快速排序，使数组有序

第一个数字的下标first的范围：[0,nums.length-3]

第二个数字的下标second的范围：[first+1, nums.length-2]

第三个数字的下标third的范围：[second+1, nums.length-1]

加入结果需满足的条件： nums[sencond]+nums[third] == - nums[first]

注意的点：结果不能重复

class Solution {
    public void quicksort(int[] nums, int l, int r){
        if(l < r){
            int p = partition(nums, l, r);
            quicksort(nums, l, p-1);
            quicksort(nums, p+1,r);
        }
    }
    public int partition(int[] nums, int start, int end){
        int base = nums[start];
        int l = start;
        int h = end;
        while(l < h){
            while(l < h && nums[h] > base){
                h--;
            }
            nums[l] = nums[h];
            while(l < h && nums[l] <= base){
                l++;
            }
            nums[h] = nums[l];
        }
        nums[l] = base;
        return l;
    }
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        int len = nums.length;
        quicksort(nums, 0, len-1);
        //Arrays.sort(nums);
        if( len < 3 || nums[0]> 0 || nums[len-1] <0){
            return res;
        }
        for(int first = 0; first <= len-3; first++ ){
            if(first > 0 && nums[first]==nums[first-1]){
                continue;
            }
            int third = len -1;
            int target = -nums[first];
            for(int second = first+1; second < len-1; second++){
                if(second > first+1 && nums[second] == nums[second-1]){
                    continue;
                }
                while(second < third && (nums[second]+nums[third])> target){
                    third--;
                }
                if(second == third){
                    break;
                }
                if((nums[second]+nums[third])== target){
                    List<Integer> route = new ArrayList<>();
                    route.add(nums[first]);
                    route.add(nums[second]);
                    route.add(nums[third]);
                    res.add(route);
                }
            }
        }
        return res;
    }
}