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;
}
}