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BJ001-1904150 Week2 homework #305
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Original file line number | Diff line number | Diff line change |
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class Solution(object): | ||
def findDuplicate(self, paths): | ||
""" | ||
:type paths: List[str] | ||
:rtype: List[List[str]] | ||
""" | ||
dd = {} | ||
for path in paths: | ||
plist = path.split(" ") | ||
prefix = plist[0]; | ||
for i in range(1, len(plist)): | ||
route = prefix + "/" + plist[i].split("(")[0] | ||
content = plist[i].split("(")[1].split(")")[0] | ||
if content not in dd: | ||
dd[content] = [] | ||
dd[content].append(route) | ||
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res = [] | ||
for key in dd: | ||
if len(dd[key]) > 1: | ||
res.append(dd[key]) | ||
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return res |
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,31 @@ | ||
/** | ||
* Definition for a binary tree node. | ||
* struct TreeNode { | ||
* int val; | ||
* TreeNode *left; | ||
* TreeNode *right; | ||
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} | ||
* }; | ||
*/ | ||
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class Solution { | ||
public: | ||
int findSecondMinimumValue(TreeNode* root) { | ||
if(root == NULL && root->left == NULL) | ||
return -1; | ||
long long ans = findmin(root, root->val); | ||
return ans == 2147483648 ? -1 : ans; | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 2147483648 是什么魔法数字?同学加点注释说说你的思路吧 |
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} | ||
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long long findmin(TreeNode* root, int val) { | ||
if(root) { | ||
if(root->val > val) { | ||
return root->val; | ||
} | ||
return min(findmin(root->left, val), findmin(root->right, val) ); | ||
} | ||
return 2147483648; | ||
} | ||
}; | ||
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|
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@@ -0,0 +1,43 @@ | ||
typedef pair<string, int> P; | ||
struct cmp | ||
{ | ||
bool operator()(const P p1, const P p2) | ||
{ | ||
if(p1.second != p2.second) | ||
return p1.second < p2.second; | ||
return p1.first > p2.first; | ||
} | ||
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}; | ||
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class Solution { | ||
public: | ||
vector<string> topKFrequent(vector<string>& words, int k) { | ||
vector<string> ans; | ||
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map<string, int> mp; | ||
map<string, int> :: iterator iter; | ||
for(int i = 0; i < words.size(); i++) { | ||
string tmp = words[i]; | ||
if(mp.find(tmp) != mp.end()) { | ||
mp[tmp]++; | ||
} else { | ||
mp[tmp] = 1; | ||
} | ||
} | ||
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priority_queue<P, vector<P>, cmp> que; | ||
for(iter = mp.begin(); iter != mp.end(); iter++) { | ||
que.push(P(iter->first, iter->second)); | ||
} | ||
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while(ans.size() != k) { | ||
P tmp = que.top(); | ||
ans.push_back(tmp.first); | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 你这没法保证频次相同的按字母顺序排呀 |
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que.pop(); | ||
} | ||
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return ans; | ||
} | ||
}; | ||
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你这方案倒是不错,能加点注释说明思路吗?