Question :
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
For example,
Given[100, 4, 200, 1, 3, 2]
,
The longest consecutive elements sequence is[1, 2, 3, 4]
. Return its length:4
.
Your algorithm should run in O(n) complexity.
Anwser 1 :
class Solution { public: int longestConsecutive(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function map<int, int> hmap; hmap.clear(); int n = num.size(); for(int i=0; i<n; i++){ //hmap.insert(pair<int, int>(num[i], i)); hmap[num[i]] = i; } int ans=0, cnt=0; map<int, int>::iterator it; for(int i=0; i<n; i++) { int cur = num[i]; it = hmap.find(num[i]); // value cnt++; if(it!=hmap.end()) { map<int, int>::iterator iter; while(1){ iter = hmap.find(++cur); // to larger value if(iter == hmap.end()) break; cnt++; hmap.erase(iter); } cur=num[i]; while(1){ iter = hmap.find(--cur); // to smaller value if(iter == hmap.end()) break; cnt++; hmap.erase(iter); } ans = cnt > ans ? cnt : ans; } cnt=0; // init to count remaider value of hmap } return ans; } };注意点:
1) 采用map,value为map_key, index为map_value
2) 按value++, value--查找,找到了则从map移除(erase),减少for循环find的次数
3) for + while,时间复杂度最坏其实为O(n*n),但是仍然编译通过了...
Anwser 2 :
class Solution { public: int getCount(map<int, int> &hmap, int value, bool asc){ int count = 0; while(hmap.find(value) != hmap.end()){ hmap.erase(value); count++; if(asc){ value++; } else { value--; } } return count; } int longestConsecutive(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function map<int, int> hmap; hmap.clear(); int n = num.size(); for(int i=0; i<n; i++){ //hmap.insert(pair<int, int>(num[i], i)); hmap[num[i]] = i; } int ans=0, cnt=0; for(int i=0; i<n; i++) { int count = getCount(hmap, num[i], false) + getCount(hmap, num[i]+1, true); ans = count > ans ? count : ans; } return ans; } };注意点:
1) 改进了方法1
2) getCount()找到则移除,count++,继续寻找下一个
3) count = getCount(hmap, num[i], false) + getCount(hmap, num[i]+1, true); 注意暗红色部分,加1为因为之前num[i]已经被移除了,因此找其上面一个value + 1
Anwser 3 :
class Solution { public: int longestConsecutive(vector<int> &num) { // Start typing your C/C++ solution below // DO NOT write int main() function map<int, int> hmap; hmap.clear(); int n = num.size(); for(int i=0; i<n; i++){ hmap.insert(pair<int, int>(num[i], 1)); // map will auto sort by key(num[i]) //hmap[num[i]] = 1; } int ans = 1; // min init int pre_k = 0; int pre_v = 0; map<int, int>::iterator it; for(it = hmap.begin(); it != hmap.end(); it++) { if(it == hmap.begin()) { pre_k = it->first; pre_v = it->second; continue; } if(it->first == pre_k + 1){ it->second = pre_v + 1; ans = it->second > ans ? it->second : ans; } pre_k = it->first; pre_v = it->second; } return ans; } };注意点:
1) 方法2的进一步改进
2) 此方法最大的利用了map自动根据key(num[i])排序,且value设为了1(起始值)