1428. Leftmost Column with at Least a One

這是一題二分搜尋法的變化題, 主要就是要把col當作二分搜尋法, 當發現有一我們就把r = mid- 1, 當沒有1就是 l = mid + 1, 直到 l > r

算是蠻新穎的題目, 可以try看看, 最後面需要檢查l 是否已經大於col數

/**
 * // This is the BinaryMatrix's API interface.
 * // You should not implement it, or speculate about its implementation
 * class BinaryMatrix {
 *   public:
 *     int get(int row, int col);
 *     vector<int> dimensions();
 * };
 */
class Solution {
public:
    bool checkcol(BinaryMatrix &b, int col){
        auto dims = b.dimensions();
        int rows = dims[0];
        for(int i = 0; i < rows;i++){
     
            if(b.get(i, col) == 1){
                return true;
            }
        }
        return false;
    }
    int leftMostColumnWithOne(BinaryMatrix &binaryMatrix) {
        auto dims = binaryMatrix.dimensions();
        int rows = dims[0];
        int cols = dims[1];
        int l = 0;
        int r = cols -1;
        while(l <= r){
            int mid =  (l + r) /2;
            if(checkcol(binaryMatrix, mid)){
                r = mid - 1;
            }else{
                l = mid + 1;
            }
        }
        if( l < cols && checkcol(binaryMatrix, l)){
            return l;
        }
        return -1;
    }
};