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combinationSum.php
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54 lines (36 loc) · 1.12 KB
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<?php
class Solution
{
public $result = [];
/**
* 给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
* candidates 中的数字可以无限制重复被选取。
* @param Integer[] $candidates
* @param Integer $target
* @return Integer[][]
*/
function combinationSum($candidates, $target)
{
$this->backTrack($candidates, $target, 0, []);
return $this->result;
}
function backTrack($candidates, $left, $step, $path)
{
if($left === 0){
$this->result[] = $path;
return;
}
if($step === count($candidates)){
return;
}
for ($i = 0; $i <= intval($left / $candidates[$step]); $i++){
for ($j = 0; $j < $i; $j++){
$path[] = $candidates[$step];
}
$this->backTrack($candidates, $left - $i * $candidates[$step], $step + 1, $path);
for ($j = 0; $j < $i; $j++){
array_pop($path);
}
}
}
}