leetcode 困难题 1520. 最多的不重叠子字符串

Problem: 1520. 最多的不重叠子字符串对每种字符得到开始和结束的区间然后对每种字符的区间内部的字符计算得到满足题意的最长区间也就是当前字符和内部字符两者区间的并集最后按照字符开始索引排序回溯得到选择的区间的总长度选择的个数个数最大化若个数相同则选择总长度更小的Codeclass Solution { public: vectorpairint, int ump, tr; vectorint tmp, ret, mem, same; int m, mxmx -1, c1, c3, ind, mi INT_MAX; void update(int cnt, int len) { if(mxmx cnt) { mxmx cnt; ret tmp; mi len; } else if (mxmx cnt) { if(mi len) ret tmp; } } int dfs( int index, int end, int cnt, int len ) { if(len mi) return INT_MIN/100; if(index m) { update(cnt, len); return 0; } // if(mem[index] 0) { // tmp.push_back(); // } int r1, r2, mx, sss, eee; for( int i index; i tr.size(); i ) { sss tr[i].first; eee tr[i].second; if( sss end ) { tmp.push_back(i); r1 dfs( i 1, eee, cnt 1, len eee - sss 1) 1; tmp.pop_back(); if(same[i] 0) r2 dfs( i 1, end, cnt, len); mx max( { mx, r1, r2 } ); } } // mem[index] mx; return mx; } vectorstring maxNumOfSubstrings(string s) { ump.assign(26, {-1, -1}); mem.assign(26, -1); int n s.size(), ch; for( int i 0; i n; i ) { ch s[i] - a; if(ump[ch].first 0) ump[ch].second ump[ch].first i; else ump[ch].second i; } int start, end, key, change 1; while(change 0) { change -1; for(int i 0; i 26; i) { if(ump[i].first 0) continue; start ump[i].first; end ump[i].second; if(end - start 1) continue; unordered_setchar te( s.begin() ump[i].first 1, s.begin() ump[i].second ); for(const char cc : te) { if(start ump[cc-a].first) { change 1; start ump[cc-a].first; } if(end ump[cc-a].second) { change 1; end ump[cc-a].second; } } ump[i] {start, end}; } } unordered_setlong long te; for(int i 0; i 26; i) { if(ump[i].first 0) continue; start ump[i].first; end ump[i].second; key (start 20) end; if(te.find(key) ! te.end()) continue; tr.push_back(ump[i]); te.insert(key); } sort(tr.begin(), tr.end()); m tr.size(); same.assign(m, -1); for(int i 0; i m; i) { if(end - start 1) { same[i] 1; continue; } unordered_setchar te( s.begin() ump[i].first, s.begin() ump[i].second); if(te.size() 1) same[i] 1; } dfs(0, -1, 0, 0); vectorstring answer; for(int i 0; i ret.size(); i) { ch ret[i]; answer.push_back( s.substr(tr[ch].first, tr[ch].second - tr[ch].first 1) ); } return answer; } };