【LeetCode】040. Combination Sum II-白红宇

【LeetCode】040. Combination Sum II

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发布时间：2019-06-12

本文共 2799 字，大约阅读时间需要 9 分钟。

题目：

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.

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]]

题解：

Solution 1 (TLE)

class Solution {public:    void dfs(vector
     
      
       >& vv, vector
       
        & v, vector
        
          candidates, int target, int sum, vector
         
          & visited) {        if(sum == target) {            vector
          
           * tmp = new vector
           
            ; *tmp = v; sort((*tmp).begin(), (*tmp).end()); if(find(vv.begin(), vv.end(), *tmp) == vv.end()) vv.push_back(*tmp); delete tmp; return; } if(sum > target) return; for(int i=0; i

> combinationSum2(vector

& candidates, int target) { vector

> vv; vector

v; vector

visited(candidates.size(),0); dfs(vv, v, candidates, target, 0, visited); return vv; }};

　　Solution 1 中即使i从level开始遍历，也无法accepted，加入stop后才勉强通过，即Solution 2

Solution 2 (almost TLE but not 579ms)

class Solution {public:    void dfs(vector
     
      
       >& vv, vector
       
        & v, vector
        
          candidates, int target, int sum, vector
         
          & visited, int stop,int level) {        if(sum == target) {            vector
          
           * tmp = new vector
           
            ; *tmp = v; sort((*tmp).begin(), (*tmp).end()); if(find(vv.begin(), vv.end(), *tmp) == vv.end()) vv.push_back(*tmp); delete tmp; return; } if(sum > target) {stop = 1;return;} for(int i=level; i

> combinationSum2(vector

& candidates, int target) { vector

> vv; vector

v; vector

visited(candidates.size(),0); sort(candidates.begin(), candidates.end()); dfs(vv, v, candidates, target, 0, visited, 0, 0); return vv; }};

Solution 3 ()

class Solution {public:    void dfs(vector
     
      
       >& vv, vector
       
        & v, vector
        
          candidates, int target, vector
         
          & visited, int stop,int level) {        if(target == 0) {            vector
          
           * tmp = new vector
           
            ; *tmp = v; sort((*tmp).begin(), (*tmp).end()); if(find(vv.begin(), vv.end(), *tmp) == vv.end()) vv.push_back(*tmp); delete tmp; return; } if(target<0) {stop = 1;return;} for(int i=level; i

> combinationSum2(vector

& candidates, int target) { vector

> vv; vector

v; vector

visited(candidates.size(),0); sort(candidates.begin(), candidates.end()); dfs(vv, v, candidates, target, visited, 0,0); return vv; }};

Solution 4

转载于:https://www.cnblogs.com/Atanisi/p/6789004.html