2 Binary Search & LogN Algorithm Apr 18

38. Search a 2D Matrix II

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

这道题与二分法有什么关系呢？

　　-把整个二维数组从对角线分成两半，从左下角开始，往右上角逼近。

public class Solution {
    /**
     * @param matrix: A list of lists of integers
     * @param target: An integer you want to search in matrix
     * @return: An integer indicate the total occurrence of target in the given matrix
     */
    public int searchMatrix(int[][] matrix, int target) {
        // write your code here
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int row = matrix.length - 1, col = matrix[0].length - 1;
        int r = row, c = 0;
        int result = 0;
        while(r >= 0 && c <= col) {
            if(matrix[r][c] == target) {
                result++;
                c++;
                r--;
            } else if(matrix[r][c] < target) {
                c++;
            } else if(matrix[r][c] > target) {
                r--;
            }
        }
        return result;
    }
}

457. Classical Binary Search

https://www.lintcode.com/problem/classical-binary-search/description?_from=ladder&&fromId=1

很简单，套用模版即可

public class Solution {
    /**
     * @param nums: An integer array sorted in ascending order
     * @param target: An integer
     * @return: An integer
     */
    public int findPosition(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) return mid;
            if(nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(nums[start] == target) {
            return start;
        } else if(nums[end] == target) {
            return end;
        } 
        return -1;
    }
}

141. Sqrt(x)

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

凡是 平方、移位，一定要类型转换！！！转换成 long ！！！

套用模版即可

public class Solution {
    /**
     * @param x: An integer
     * @return: The sqrt of x
     */
    public int sqrt(int x) {
        // write your code here
        if(x < 0) return -1;
        long start = 0; 
        long end = x;
        long xl = (long)x;
        while(start + 1 < end) {
            long mid = start + (end - start) / 2;
            if(mid * mid == xl) {
                return (int)(mid);
            } else if(mid * mid > xl) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if(end * end <= xl) {
            return (int)(end);
        } else {
            return (int)(start);
        }
    }
}

617.  Maximum Average Subarray

https://www.lintcode.com/problem/maximum-average-subarray-ii/note/171478

586. Sqrt(x) II

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

牛顿迭代法。

public class Solution {
    /**
     * @param x: a double
     * @return: the square root of x
     */
    public double sqrt(double x) {
        // write your code here
        double result = 1.0;
        while(Math.abs(x - result * result) > 1e-12) {
            result = (result + x / result) / 2;
        }
        return result;
    }
}

160. Find Minimum in Rotated Array II

https://www.lintcode.com/problem/find-minimum-in-rotated-sorted-array-ii/description?_from=ladder&&fromId=1

要知道最坏情况下，[1, 1, 1, .........., 1] 里有一个 0

这种情况使得时间复杂度必须是 O(n)

if mid equals to end, that means it's fine to remove end

the smallest element won't be removed

public class Solution {
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        // write your code here
        if(nums == null || nums.length == 0) return 0;
        int start = 0;
        int end = nums.length - 1;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(nums[mid] == nums[end]) {
                end--;
            }
            if(nums[mid] > nums[end]) {
                start = mid;
            }
            if(nums[mid] < nums[end]) {
                end = mid;
            }
        }
        return Math.min(nums[start], nums[end]);
    }
}

63. Search in Rotated Sorted Array II

https://www.lintcode.com/problem/search-in-rotated-sorted-array-ii/description?_from=ladder&&fromId=1

class Solution {
    public boolean search(int[] nums, int target) {
        if(nums == null || nums.length == 0) return false;
        int start = 0;
        int end = nums.length - 1;
        int mid;
        while(start + 1 < end){
            mid = start + (end - start) / 2;
            if(nums[mid] == target) return true;
            if(nums[mid] > nums[start]){
                if(target >= nums[start] && target <= nums[mid]){
                    end = mid;
                }else{
                    start = mid;
                }
            }else if(nums[mid] < nums[start]){
                if(target >= nums[mid] && target <= nums[end]){
                    start = mid;
                }else{
                    end = mid;
                }
            }
            else{
                start++;
            }
        }
        if(target == nums[start] || nums[end] == target) return true;
        return false;
    }
}