leetcode topic

Combined total -- leetcode 39

Title Description

Given an array candidates without repeating elements and a target number target,
Find out all the combinations of numbers and target s in candidates.
The numbers in candidates can be selected repeatedly without limit.

Explain:
All numbers, including target, are positive integers.
A solution set cannot contain duplicate combinations. 

Example 1:
Input: candidates = [2,3,6,7], target = 7,
The solution set is:
[
  [7],
  [2,2,3]
]

Example 2:
Input: candidates = [2,3,5], target = 8,
The solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

Source: LeetCode
 Link: https://leetcode-cn.com/problems/combination-sum
 Copyright belongs to the network. For commercial reprint, please contact the official authorization. For non-commercial reprint, please indicate the source.

thinking

Backtracking algorithm
 Recursively find and combine target, exit and exceed target

Code

package com.leetcode.array;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * Title: 
 * Combined total -- leetcode 39
 * 
 * Title Description:
 * 
Given an array candidates without repeating elements and a target number target,
Find out all the combinations of numbers and target s in candidates.
candidates The number in can be selected repeatedly without limit.

Explain:
All numbers, including target, are positive integers.
A solution set cannot contain duplicate combinations. 

Example 1:
Input: candidates = [2,3,6,7], target = 7,
The solution set is:
[
  [7],
  [2,2,3]
]

Example 2:
Input: candidates = [2,3,5], target = 8,
The solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

Source: LeetCode
 Link: https://leetcode-cn.com/problems/combination-sum
 Copyright belongs to the network. For commercial reprint, please contact the official authorization. For non-commercial reprint, please indicate the source.
 */
public class CombinationSum {
	/**
	 * Train of thought: 
	 * 1,Backtracking algorithm
	 * 2,Recursively find and combine target, exit and exceed target
	 */
    public List<List<Integer>> combinationSum(int[] bb, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if (bb == null) {
            return res;
        }
        
        addCombinations(bb, 0, target, new ArrayList<Integer>(), res);
        
        return res;
    }
    
    private void addCombinations(int[] bb, int start, int target, List<Integer> cache, List<List<Integer>> res) {
        if (target < 0) {
            return;
        }
        if (target == 0) {
            res.add(new ArrayList<>(cache));
            return;
        }
        for (int i=start; i<bb.length; i++) {
            cache.add(bb[i]);
            addCombinations(bb,i,target-bb[i],cache,res);
            cache.remove(cache.size()-1);
        }
    }
	
    
    /**
     * Train of thought: 
     * Backtracking after optimization
     */
    public List<List<Integer>> combinationSumII(int[] bb, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if (bb == null) {
            return res;
        }
        
        // After sorting the array, the number of recursions can be reduced during recursion, and if (BB [i] > target) break can be used.
        Arrays.sort(bb);
        
        addCombinationsII(bb, 0, target, new ArrayList<Integer>(), res);
        
        return res;
    }
    
    private void addCombinationsII(int[] bb, int start, int target, List<Integer> cache, List<List<Integer>> res) {
        if (target < 0) {
            return;
        }
        if (target == 0) {
            res.add(new ArrayList<>(cache));
            return;
        }
        for (int i=start; i<bb.length; i++) {
        	// Improve performance with sorted array
            if (bb[i] > target) {
                break;
            }
            cache.add(bb[i]);
            addCombinationsII(bb,i,target-bb[i],cache,res);
            cache.remove(cache.size()-1);
        }
    }
}