leetcode18. 4Sum

题目

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.A solution set is:[  [-1,  0, 0, 1],  [-2, -1, 1, 2],  [-2,  0, 0, 2]]

思路

在原来3sum的基础上面加一层循环

代码

public class Solution {    public List<List<Integer>> fourSum(int[] nums, int target) {        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();        if(nums.length < 4){            return res;        }        Arrays.sort(nums);        int len = nums.length;        int max = nums[len - 1];        if(nums[0] * 4  > target  || max * 4 < target){            return res;        }        int t;        for( int i = 0; i < len - 3; i ++){            t = nums[i];            if(i > 0 && t == nums[i - 1])                continue;            if(4 * t > target)                break;            if(4 * t == target && i < len - 3&& nums[i + 3] == t){                res.add(Arrays.asList(t, t, t, t));                continue;            }               threeSum(nums, target - t, i+1, len, res, t);        }        return res;    }    public void  threeSum(int[] nums, int target, int lo, int hi, List<List<Integer>> res, int t) { for(int i  = lo; i <  hi- 2; i++){            if(i == lo || ( nums[i] != nums[i-1])){                int low = i+1;int high = nums.length - 1; int iSum = target  - nums[i];                while(low < high){                    if(nums[low] + nums[high] == iSum){                        res.add(Arrays.asList(t, nums[i], nums[low], nums[high]));                        while(low < high && nums[low] == nums[low + 1]) low++;                        while(low < high && nums[high] == nums[high - 1]) high--;                        low++; high--;                    }                    else if( nums[low] + nums[high] > iSum){                        high--;                    }                    else{                        low++;                    }                }            }        }    }}

他山之玉

public List<List<Integer>> fourSum(int[] nums, int target) {        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();        int len = nums.length;        if (nums == null || len < 4)            return res;        Arrays.sort(nums);        int max = nums[len - 1];        if (4 * nums[0] > target || 4 * max < target)            return res;        int i, z;        for (i = 0; i < len; i++) {            z = nums[i];            if (i > 0 && z == nums[i - 1])// avoid duplicate                continue;            if (z + 3 * max < target) // z is too small                continue;            if (4 * z > target) // z is too large                break;            if (4 * z == target) { // z is the boundary                if (i + 3 < len && nums[i + 3] == z)                    res.add(Arrays.asList(z, z, z, z));                break;            }            threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);        }        return res;    }    /*     * Find all possible distinguished three numbers adding up to the target     * in sorted array nums[] between indices low and high. If there are,     * add all of them into the ArrayList fourSumList, using     * fourSumList.add(Arrays.asList(z1, the three numbers))     */    public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,            int z1) {        if (low + 1 >= high)            return;        int max = nums[high];        if (3 * nums[low] > target || 3 * max < target)            return;        int i, z;        for (i = low; i < high - 1; i++) {            z = nums[i];            if (i > low && z == nums[i - 1]) // avoid duplicate                continue;            if (z + 2 * max < target) // z is too small                continue;            if (3 * z > target) // z is too large                break;            if (3 * z == target) { // z is the boundary                if (i + 1 < high && nums[i + 2] == z)                    fourSumList.add(Arrays.asList(z1, z, z, z));                break;            }            twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);        }    }    /*     * Find all possible distinguished two numbers adding up to the target     * in sorted array nums[] between indices low and high. If there are,     * add all of them into the ArrayList fourSumList, using     * fourSumList.add(Arrays.asList(z1, z2, the two numbers))     */    public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,            int z1, int z2) {        if (low >= high)            return;        if (2 * nums[low] > target || 2 * nums[high] < target)            return;        int i = low, j = high, sum, x;        while (i < j) {            sum = nums[i] + nums[j];            if (sum == target) {                fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));                x = nums[i];                while (++i < j && x == nums[i]) // avoid duplicate                    ;                x = nums[j];                while (i < --j && x == nums[j]) // avoid duplicate                    ;            }            if (sum < target)                i++;            if (sum > target)                j--;        }        return;    }