Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC".
Note:
If there is no such window in S that covers all characters in T, return the emtpy string "".
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
给定数字n和k, 返回所有由1,2,3...,n数组成k元组合
递归
class Solution {
public:
void dfs(vector<vector<int> >&result, vector<int>combination, int kth, int k, int curVal, int n){
// combination 已经生成的前kth-1个数
// kth 本次将要生成组合中的第kth个数
// k 组合规模
// curVal 本次将加入组合中的数，即第kth个数
// n 候选数的总规模
combination.push_back(curVal);
if(kth==k)result.push_back(combination);
else{
for(int val=curVal+1; val+(k-kth-1)<=n; val++){
dfs(result, combination, kth+1, k, val, n);
}
}
}
vector<vector<int> > combine(int n, int k) {
vector<vector<int> >result;
if(k>n || n<0 || k<1)return result;
vector<int> combination;
for(int val=1; val<=n-(k-1); val++){
dfs(result, combination, 1, k, val, n);
}
return result;
}
};LeetCode: Combinations [077],布布扣,bubuko.com
原文地址：http://blog.csdn.net/harryhuang1990/article/details/27522955
