• to do

**1] Set Matrix Zeroes **

• naive 36%

   void setZeroes(vector<vector<int>>& matrix) {
        if (!matrix.size() || !matrix[0].size()) return;
        vector<bool> markedRow(matrix.size(), false);
        vector<bool> markedCol(matrix[0].size(), false);

        for (int i=0; i<markedRow.size(); ++i) {
            for (int j=0; j<markedCol.size(); ++j) {
                if (matrix[i][j]==0) {
                    markedRow[i] = markedCol[j] = true;
                }
            }
        }

        for (int i=0; i<markedRow.size(); ++i) {
            if (markedRow[i]) {
                fill(matrix[i].begin(), matrix[i].end(), 0);
            }
        }

        for (int j=0; j<markedCol.size(); ++j) {
            if (markedCol[j]) {
                for (int i=0; i<matrix.size(); ++i) {
                    matrix[i][j] = 0;
                } 
            }
        }           
    }

**2] Gas station **
timeout, try dp -> 最大子序列和

    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        for (int i = 0; i < gas.size(); ++i) {
            // try starting at station i, record at curr station j
            int j = i;
            int curr = gas[j];
            while (curr >= cost[j]) { // still enough to reach next station
                curr -= cost[j]; // reach next station
                j = (j + 1) % gas.size();
                if (j == i) return i;
                else curr += gas[j];
            }
        }
        return -1;
    }