leetcode14 4sum

題目要求

此處爲原題地址

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]
]

也就是從數組中找到全部四個數字，這四個數字的和爲目標值。這四個數字的組合不能重複。

這題的核心思路請參考個人另外一篇博客three-sum，即如何找到三個數字，使其和爲目標值。

思路一：直接利用three-sum

在three-sum的基礎上在外圍再加一圈循環，即在最左側數字固定的狀況下，尋找右側數組中的three-sum結果。時間複雜度O(n3)。這裏須要注意及時處理掉重複的狀況。

public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        int length = nums.length;
        if(length<4){
            return result;
        }
        Arrays.sort(nums);
        for(int i = 0 ; i < length-3 ; ){
            int firstValue = nums[i];
            
            //three sum
            for(int j = i+1 ; j < length-2 ; ){
                int secondValue = nums[j];
                int leftPointer = j+1;
                int rightPointer = length-1;
                while(leftPointer<rightPointer){
                    int currentSum = firstValue + secondValue + nums[leftPointer] + nums[rightPointer];
                    if(currentSum == target){
                        result.add(Arrays.asList(firstValue, secondValue, nums[leftPointer], nums[rightPointer]));
                    }
                    if(currentSum<=target){
                        while(nums[leftPointer]==nums[++leftPointer] && leftPointer<rightPointer);
                    }
                    if(currentSum>=target){
                        while(nums[rightPointer--]==nums[rightPointer] && leftPointer<rightPointer);
                    }
                }
                //去除重複狀況
                while(nums[j] == nums[++j] && j < length - 2);
            }
            //去除重複狀況
            while(nums[i] == nums[++i] && i < length - 3);
        }
        return result;
    }

提升算法的效率

在three-sum的基礎上計算4sum不可避免的須要O(n3)的時間複雜度。那麼就須要儘量排除不可能的狀況來提升計算效率。由於數組已經被排序，因此能夠根據數組中元素的位置判斷接下來的狀況是否有可能合成目標值。

//思路二：不斷的排除不可能的狀況，以加快遍歷，核心的仍是3sum
    public List<List<Integer>> fourSum2(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 防止重複
                continue;
            if (z + 3 * max < target) // z is too small 當前值即時加上最大值也不足以構成目標值，進入下一輪循環
                continue;
            if (4 * z > target) // z is too large 當前值*4都大於目標值，餘下的值不可能再生成最大值
                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;
    }

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