lintCode二分查找

1、x 的平方根

class Solution:
    """
    @param x: An integer
    @return: The sqrt of x
    """
    def sqrt(self, x):
        # write your code here
        left, right = 0, 100000
        while left < right:
            mid = (left + right) // 2
            if mid * mid == x:
                return mid
            elif mid * mid < x:
                left = mid + 1
            else:
                right = mid - 1
        if left * left > x:
            return left-1
        return left

2、搜索插入位置

class Solution:
    """
    @param A: an integer sorted array
    @param target: an integer to be inserted
    @return: An integer
    """
    def searchInsert(self, A, target):
        # write your code here
        if not A: return 0
        left, right = 0, len(A) - 1
        while left != right:
            mid = (left + right) // 2
            if A[mid] == target:
                return mid
            elif A[mid] < target:
                left = mid +1 
            else:
                right = mid
        if A[left] < target:
            return left + 1
        else:
            return left

3、搜索二维矩阵

class Solution:
    """
    @param matrix: matrix, a list of lists of integers
    @param target: An integer
    @return: a boolean, indicate whether matrix contains target
    """
    def searchMatrix(self, matrix, target):
        # write your code here
        if not matrix:return False
        m = len(matrix)
        n = len(matrix[0])
        left, right = 0, m * n - 1
        while left <= right:
            mid = (left + right) // 2
            if matrix[mid // n][mid % n] == target:
                return True
            elif matrix[mid // n][mid % n] < target:
                left = mid + 1
            else:
                right -= 1
        return False

4、二分查找

class Solution:
    """
    @param nums: The integer array.
    @param target: Target to find.
    @return: The first position of target. Position starts from 0.
    """
    def binarySearch(self, nums, target):
        # write your code here
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                if mid == 0 or nums[mid - 1] != target:
                    return mid
                else:
                    right = mid
            elif nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return -1

5、寻找旋转排序数组中的最小值

思路：一样左右向中间用二分法，每次都保证左边数值大于右边的数值，当没法保证的时候，就是边界了。
class Solution:
    """
    @param nums: a rotated sorted array
    @return: the minimum number in the array
    """
    def findMin(self, nums):
        # write your code here
        if len(nums) == 1: return nums[0]
        left, right = 0, len(nums) - 1
        while right - left > 1:
            if nums[left] < nums[right]: return nums[left]
            mid = (left + right) // 2
            if nums[mid] > nums[left]:
                left = mid
            else:
                right = mid
        return min(nums[left], nums[right])

6、木材加工

思路：二分法想不出来，哎。只用用暴力解法。
class Solution:
    """
    @param L: Given n pieces of wood with length L[i]
    @param k: An integer
    @return: The maximum length of the small pieces
    """
    def woodCut(self, l, k):
        # write your code here
        if not l:return 0
        max_len = sum(l) // k
        while max_len>0:
            now_k = 0
            for i in l:
                now_k += i // max_len
            if now_k >= k:
                return max_len
            max_len -= 1
        return 0