leetcode-698-Partition to K Equal Sum Subsets
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问题
题目:[leetcode-698]
思路
能想到的是搜索的思路,但是不知道怎么用。
参考了[698. Partition to K Equal Sum Subsets]
看了别人的代码,知道这个题搜索的技巧在于,对于每一个元素nums[i],枚举每一个子数组的当前位置。这也是一种枚举策略。并且,这个题目只有这么理解枚举,才能做出来。
代码(wa)
下面枚举的思路没有错,就是这个思路。但是TLE。
class Solution {public: bool canPartitionKSubsets(vector<int>& nums, int k) { int sz = nums.size(); int sum = 0; for(int i = 0; i < sz; ++i) { sum += nums[i]; } if( sum % k != 0 ) return false; else{ vector<int> ret(k , 0); bool flag = false; dfs( 0, nums, sum/k, ret, flag ); return flag; } }private: void dfs( int depth, vector<int>& nums, int target, vector<int>& ret, bool& flag ) { if( flag ) return; if( depth == nums.size() ) { int k = ret.size(); for(int i = 0; i < k; ++i) { if( ret[i] != target ) { flag = false; return; } } flag = true; } else { int k = ret.size(); for( int i = 0; i < k; ++i ) { if( ret[i] + nums[depth] <= target ) { ret[i] += nums[depth]; dfs( depth + 1, nums, target, ret, flag ); ret[i] -= nums[depth]; } } } }};
代码1(ac)
class Solution {public: bool canPartitionKSubsets(vector<int>& nums, int k) { int sz = nums.size(); int sum = 0; for(int i = 0; i < sz; ++i) { sum += nums[i]; } if( sum % k != 0 ) return false; for( int i = 0; i < sz; ++i ) if( nums[i] > sum/k ) return false; /* 排序后倒置数组,可以ac。应该是刚好能过,看建议的方法要记忆化搜索了。下面的方法并没有记忆化搜索 */ sort( nums.begin(), nums.end() ); vector<int> new_nums; for( int i = 0; i < sz; ++i ) new_nums.push_back(nums[sz-1-i] ); vector<int> ret(k , 0); bool flag = false; dfs( 0, new_nums, sum/k, ret, flag ); return flag; }private: void dfs( int depth, vector<int>& nums, int target, vector<int>& ret, bool& flag ) { if( flag ) return; if( depth == nums.size() ) { int k = ret.size(); for(int i = 0; i < k; ++i) { if( ret[i] != target ) { flag = false; return; } } flag = true; } else { int k = ret.size(); for( int i = 0; i < k; ++i ) { if( ret[i] + nums[depth] <= target ) { ret[i] += nums[depth]; dfs( depth + 1, nums, target, ret, flag ); ret[i] -= nums[depth]; } } } }};
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