How to calculate combined sum using backtracking algorithm in PHP

Given an array of candidates and a target number target, find all the combinations in candidates that can make the sum of the numbers be target. What should we do at this time? Today I will take you through it.

Given an array candidates and a target number target, find all the combinations in candidates that can make the sum of the numbers target. Each number in

candidates can only be used once in each combination.

Note:

All numbers (including the target number) are positive integers. The solution set cannot contain duplicate combinations.

Example 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
所求解集为:[
 [1, 7],
 [1, 2, 5],
 [2, 6],
 [1, 1, 6]]

Example 2:

输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:[  
[1,2,2],  
[5]]

Solution ideas

Direct reference to the backtracking algorithm group elimination permutation/combination/subset problem

Code

class Solution {
    /** * @param Integer[] $candidates * @param Integer $target * @return Integer[][] */
    public $res = [];
    function combinationSum2($candidates, $target) {
        sort($candidates);   // 排序
        $this->dfs([], $candidates, $target, 0);
        return $this->res;
    }
    function dfs($array, $candidates, $target, $start) {
        if ($target < 0) return;
        if ($target === 0) {
            $this->res[] = $array;
            return;
        }
        $count = count($candidates);
        for ($i = $start; $i < $count; $i++) {
            if ($i !== $start && $candidates[$i] === $candidates[$i - 1]) continue;
            $array[] = $candidates[$i];
            $this->dfs($array, $candidates, $target - $candidates[$i], $i + 1);//数字不能重复使用,需要+1
            array_pop($array);
        }
    }}

Extra:

Given a non-repeating element An array of candidates and a target number target, find all the combinations in candidates that can make the sum of numbers be target. The numbers in

candidates can be selected repeatedly without limit.

The difference is that repeated selections are allowed, and it is solved by making two changes based on the previous question.

class Solution {
    /** * @param Integer[] $candidates * @param Integer $target * @return Integer[][] */
    public $res = [];
    function combinationSum($candidates, $target) {
        sort($candidates);   // 排序
        $this->dfs([], $candidates, $target, 0);
        return $this->res;
    }
    function dfs($array, $candidates, $target, $start) {
        if ($target < 0) return;
        if ($target === 0) {
            $this->res[] = $array;
            return;
        }
        $count = count($candidates);
        for ($i = $start; $i < $count; $i++) {
            // if ($i !== $start && $candidates[$i] === $candidates[$i - 1]) continue; // 注释掉去重的代码
            $array[] = $candidates[$i];
            $this->dfs($array, $candidates, $target - $candidates[$i], $i);//数字能重复使用, 不需要+1
            array_pop($array);
        }
    }}

Extra:

Find all k combinations of numbers whose sum is n. Only positive integers from 1 to 9 are allowed in the combination, and there are no duplicate numbers in each combination.

Limit the number of elements in the selected scheme

class Solution {
    public $res = [];
    /** * @param Integer $k * @param Integer $n * @return Integer[][] */
    function combinationSum3($k, $n) {
        $this->dfs([], [1,2,3,4,5,6,7,8,9], $n, 0, $k);
        return $this->res;
    }
    function dfs($array, $candidates, $n, $start, $k) {
        if ($n < 0) return;
        if ($n === 0 && count($array) === $k) {
            $this->res[] = $array;
            return;
        }
        for ($i = $start; $i < 9; $i++) {
            if ($i !== $start && $candidates[$i] === $candidates[$i - 1]) continue;
            $array[] = $candidates[$i];
            $this->dfs($array, $candidates, $n - $candidates[$i], $i + 1, $k);
            array_pop($array);
        }
    }}

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