【leetcode】Array——4Sum（18）

題目：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:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• 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)

自己的思路：暴力循環破解，三個指針挨個循環，第四個數用binary search查找。（這樣比用4個指針循環要好）

代碼：

public List<List<Integer>> fourSum(int[] nums, int target) {
	Arrays.sort(nums);
	
	List<List<Integer>> result = new ArrayList<List<Integer>>();
	List<Integer> cur = new ArrayList<Integer>();
	if(nums.length<4)
		return result;
	for(int i=0;i<nums.length-3;i++){
		if(i>0&&nums[i]==nums[i-1])
			continue;
		for(int j=i+1;j<nums.length-2;j++){
			if(j>i+1&&nums[j]==nums[j-1])
				continue;
			for(int k=j+1;k<nums.length-1;k++){
				if(k>j+1&&nums[k]==nums[k-1])
					continue;
				int rest = target-nums[i]-nums[j]-nums[k];
				int restIndex = binarySearch(nums, k+1, rest);
				if(restIndex!=-1){
					cur.add(nums[i]);
					cur.add(nums[j]);
					cur.add(nums[k]);
					cur.add(nums[restIndex]);
					result.add(new ArrayList<Integer>(cur));
					cur.clear();
				}
					
			}
		}
	}
	System.out.println(result);
	return result;
}

private int binarySearch(int[]nums,int start,int rest){
	int end = nums.length-1;
	while(start<=end){
		int mid = start+(end-start)/2;
		if(nums[mid]==rest)
			return mid;
		else if(nums[mid]>rest)
			end = mid -1;
		else 
			start=mid+1;
	}
	return -1;
}

代碼雖然被leetcode Accepte了，但是耗時太長。看看下面大神的解法吧！

大神的思路：先針對一個數nums[i]判斷該數是否可能，如果nums[i]+max*3<target 或則nums[i]*4>target 等其他異常情況，則直接跳過，變成3Sum問題；同樣可以化解爲2Sum問題。 算法的時間複雜度O(n^3).

鏈接：https://leetcode.com/discuss/69517/7ms-java-code-win-over-100%25

代碼：

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;
    }

另一個思路：用hashtable

map中key爲2數之和，value爲兩數的座標對的列表。

用兩層循環遍歷前兩個數，後兩個數從map中查詢

代碼：略