2 - Binary Search & LogN Algorithm

254. Drop Eggs

https://www.lintcode.com/problem/drop-eggs/description?_from=ladder&&fromId=1

28. Search a 2D Matrix

https://www.lintcode.com/problem/search-a-2d-matrix/description?_from=ladder&&fromId=1

思路1：

　　1. find the row index, the last number <= target

　　2. find the column index, the number equal to target

public class Solution {
    /**
     * @param matrix: matrix, a list of lists of integers
     * @param target: An integer
     * @return: a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        // write your code here
        if(matrix == null || matrix.length == 0) return false;
        int left = 0, right = matrix[0].length - 1;
        int row = findRow(matrix, target);
        if(matrix[row][0] > target || matrix[row][right] < target) {
            return false;
        }
        while(left + 1 < right) {
            int mid = left + (right - left) / 2;
            if(matrix[row][mid] == target) {
                return true;
            } else if(matrix[row][mid] > target) {
                right = mid;
            } else {
                left = mid;
            }
        }
        if(matrix[row][left] == target || matrix[row][right] == target) {
            return true;
        } else {
            return false;
        }
    }

    public int findRow(int[][] matrix, int target) {
        int top = 0, bottom = matrix.length - 1, right = matrix[0].length - 1;
        while(top + 1 < bottom) {
            int mid = top + (bottom - top) / 2;
            if(matrix[mid][right] == target) {
                return mid;
            } else if(matrix[mid][right] > target) {
                bottom = mid;
            } else {
                top = mid;
            }
        }
        if(matrix[top][right] >= target) {
            return top;
        } else {
            return bottom;
        }
    }
}

思路2:

1. 可以看作是一个有序数组被分成了n段，每段就是一行。因此依然可以二分求解。

对每个数字，根据其下标 i，j 进行编号，每个数字可被编号为 0 ～ n *（n - 1）

2. 相当于是在一个数组中的下标，然后直接像在数组中二分一样来做。取得 mid 要还原成二维数组中的下标，i = mid / n, j = mid % n

3. int start = 0, end = row * row * column - 1;

int number = matrix[mid / column][mid % column];

// Binary Search Once
public class Solution {
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        // write your code here
        if(matrix == null || matrix.length == 0){
            return false;
        }

        if(matrix[0] == null || matrix[0].length == 0){
            return false;
        }

        int row = matrix.length;
        int column = matrix[0].length;

        int start = 0, end = row * column - 1;
        while(start <= end){
            int mid = start + (end - start) / 2;
            int number = matrix[mid / column][mid % column];
            if(number == target){
                return true;
            }else if(number > target){
                end = mid - 1;
            }else{
                start = mid + 1;
            }
        }

        return false;

    }
}

14. First Position of Target 

https://www.lintcode.com/problem/first-position-of-target/description?_from=ladder&&fromId=1

思路：这道题很简单。套用二分法的模版即可

 public class Solution {
     /**
      * @param nums: The integer array.
      * @param target: Target to find.
      * @return: The first position of target. Position starts from 0.
      */
     public int binarySearch(int[] nums, int target) {
         // write your code here
         if(nums == null || nums.length == 0) return -1;
         int start = 0, end = nums.length - 1;
         while(start + 1 < end) {
             int mid = start + (end - start) / 2;
             if(nums[mid] >= target) {
                 end = mid;
             } else {
                 start = mid;
             }
         }
         if(nums[start] == target) {
             return start;
         }
         if(nums[end] ==  target) {
             return end;
         }
         return -1;
     }
 }

414. Divide Two Integers

https://www.lintcode.com/problem/divide-two-integers/description?_from=ladder&&fromId=1

1. 凡是要移位，要做的第一件事就是把 int 转换成 long，为了防止移位时溢出。

2. 基本思路是利用减法，看看被除数可以减去多少次除数。使用倍增的思想优化，可以将减法的次数优化到对数的时间复杂度。

3. 我们将除数左移一位（或加上它自己），即得到了二倍的除数，这时一次相当于减去了两个除数，通过不断倍增，时间复杂度很优秀。

4. 与此同时，还需要一个变量记录此时的除数是最初除数的多少倍，每次减法后都加到结果上即可。

 public class Solution {
     /**
      * @param dividend: the dividend
      * @param divisor: the divisor
      * @return: the result
      */
     public int divide(int dividend, int divisor) {
         // write your code here
         if(divisor == 0) {
             return dividend >= 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
         }
         if(dividend == 0) {
             return 0;
         }
         if(divisor == -1 && dividend == Integer.MIN_VALUE) {
             return Integer.MAX_VALUE;
         }
         boolean isNegative = ((divisor > 0 && dividend < 0) || (divisor < 0 && dividend > 0)) ? true : false;
         long divisorL = Math.abs((long)divisor);
         long dividendL = Math.abs((long)dividend);
         int result = 0;
         while(dividendL >= divisorL) {
             int shift = 0;
             while(dividendL >= (divisorL << shift)) {
                 shift++;
             }
             result += 1 << (shift - 1);
             dividendL -= divisorL << (shift - 1);
         }
         if(isNegative) {
             return result * (-1);
         }
         return result;
     }
 }

61. Search for a Range

https://www.lintcode.com/problem/search-for-a-range/description?_from=ladder&&fromId=1

这道题同样很简单，套用模版即可。

 public class Solution {
     /**
      * @param A: an integer sorted array
      * @param target: an integer to be inserted
      * @return: a list of length 2, [index1, index2]
      */
     public int[] searchRange(int[] A, int target) {
         // write your code here
         if(A == null || A.length == 0) return new int[]{-1, -1};
         int start = findStart(A, target);
         int end = findEnd(A, target);
         return new int[]{start, end};

     }

     public int findStart(int[] A, int target) {
         int start = 0;
         int end = A.length - 1;
         while(start + 1 < end) {
             int mid = start + (end - start) / 2;
             if(A[mid] < target) {
                 start = mid;
             } else {
                 end = mid;
             }
         }
         if(A[start] == target) {
             return start;
         }
         if(A[end] == target) {
             return end;
         }
         return -1;
     }

     public int findEnd(int[] A, int target) {
         int start = 0;
         int end = A.length - 1;
         while(start + 1 < end) {
             int mid = start + (end - start) / 2;
             if(A[mid] <= target) {
                 start = mid;
             } else {
                 end = mid;
             }
         }
         if(A[end] == target) {
             return end;
         }
         if(A[start] == target) {
             return start;
         }
         return -1;
     }
 }