LeetCode—数组(3)

1.Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7
A solution set is: 
[7] 
[2, 2, 3] 

 public List<List<Integer>> combinationSum(int[] candidates, int target) {
         List<List<Integer>> res=new ArrayList<List<Integer>>();
      if(candidates==null||candidates.length==0)
        	return res;
        
        Arrays.sort(candidates);
        combinationSum(candidates, target,0,new ArrayList<Integer>(),res);
        return res;
        	
   }

    private static void combinationSum(int[] candidates, int target, int start, ArrayList<Integer> list,
			List<List<Integer>> res) {
		if(target==0){
			res.add(list);
			return;
		}
		for(int i=start;i<candidates.length;i++){
			if(target-candidates[i]>=0){
				ArrayList<Integer> tmp=(ArrayList<Integer>)list.clone();
				int tmp_target=target-candidates[i];
				tmp.add(candidates[i]);
				combinationSum(candidates, tmp_target, i, tmp, res);
			}
		}
		
	}</span>

2.

Combination Sum II

 

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
          List<List<Integer>> res=new ArrayList<List<Integer>>();
        if(candidates==null||candidates.length==0)
        	return res;
        Arrays.sort(candidates);
        combinationSum2(candidates,target,0,new ArrayList<Integer>(),res);
        return res;
    }
	
	private static void combinationSum2(int[] candidates, int target, int start, ArrayList<Integer> arrayList,
			List<List<Integer>> res) {
		if(target==0){
			
			res.add(arrayList);
			return;
		}
		for(int i=start;i<candidates.length;i++){
			if (i > start && candidates[i] == candidates[i-1]) continue;
			if(target-candidates[i]>=0){
				ArrayList<Integer> tmp=(ArrayList<Integer>)arrayList.clone();
				int tmp_target=target-candidates[i];
				tmp.add(candidates[i]);
				combinationSum2(candidates, tmp_target, i+1, tmp, res);
			}
		}
		
	}</span>

Combination Sum III

 

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.


Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]


Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> res = new ArrayList<>();
        Deque<Integer> tmp = new ArrayDeque<>();
        if (n == 0 || k == 0 || n / k > 9) {
            return res;
        }
        helper(res, tmp, 1, k, n);
        return res;
    }

    private void helper(List<List<Integer>> res, Deque<Integer> tmp, int start, int k, int n) {
        if (0 == n && k == 0) {
            res.add(new ArrayList<>(tmp));
            return;
        }
        for (int i = start; i <= 9; i++) {
            tmp.addLast(i);
            helper(res, tmp, i + 1, k - 1, n - i);
            tmp.removeLast();
        }
    }</span>



### LeetCode 数组题目及解法 #### 二分查找 在处理有序数组时,二分查找是一种高效的算法。通过不断将搜索范围减半来快速定位目标值的位置[^1]。 ```cpp int binarySearch(int arr[], int l, int r, int x) { while (l <= r) { int m = l + (r - l) / 2; // Check if x is present at mid if (arr[m] == x) return m; // If x greater, ignore left half if (arr[m] < x) l = m + 1; // If x is smaller, ignore right half else r = m - 1; } // if we reach here, then element was not present return -1; } ``` #### 移除元素 针对移除指定元素的任务,可以采用双指针方法提高效率。这种方法不仅减少了不必要的遍历次数,还简化了代码逻辑[^3]。 ```cpp class Solution { public: int removeElement(vector<int>& nums, int val) { int k = 0; for (int i = 0; i < nums.size(); ++i) { if (nums[i] != val) { nums[k++] = nums[i]; } } return k; } }; ``` #### 查找重复数字 当面对只读且不允许额外空间开销的情况下寻找重复项的问题时,可以通过巧妙利用原数组特性实现线性时间内解决问题的方法[^2]。 ```cpp class Solution { public: int findDuplicate(const vector<int>& nums) const { int slow = nums[0], fast = nums[nums[0]]; while(slow != fast){ slow = nums[slow]; fast = nums[nums[fast]]; } fast = 0; while(fast != slow){ slow = nums[slow]; fast = nums[fast]; } return slow; } }; ``` 这些例子展示了如何运用不同策略解决LeetCode上的典型数组类问题。每种解决方案都考虑到了特定条件下的最优实践,旨在帮助开发者提升编程技巧并加深对数据结构的理解。
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