leetcode 【 Search Insert Position 】python 实现

题目：

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

代码：oj测试通过 Runtime: 52 ms

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be inserted
    # @return integer
    def search(self, A, start, end, target):
        # search stop case I
        if start == end:
            if start == 0:
                return [ 0,1 ][ A[0]<target ]
            if A[start] == target:
                return start
            else:
                return [start,start+1][A[start]<target]
        # search stop case II
        if start+1 == end:
            if A[start] >= target:
                return start
            elif A[start] < target and A[end] >= target :
                return end
            else:
                return end+1
        
        mid = (start+end)/2
        # search stop case III
        if A[mid] == target:
            return mid
        if A[mid] > target:
            return self.search(A, start, mid-1, target)
        if A[mid] < target:
            return self.search(A, mid+1, end, target)
        
    def searchInsert(self, A, target):
        # zero length case
        if len(A) == 0:
            return 0
        # binary search
        start = 0
        end = len(A)-1
        return self.search(A, start, end, target)

思路：

二分查找经典题。

采用迭代方式时：

1. 考虑start==end的情况（一个元素）和start+1==end的情况（两个元素），作为迭代终止的两种case。

2. 当元素数量大于3时作为一般的case处理，二分查找。

3. 根据题意要求进行判断条件。

4. 第一次提交没有AC ，原因是在处理start==end的case时候，竟然只考虑了0和len(A)的边界情况，没有考虑一般情况，陷入了思维的陷阱。

后面又写了一版非递归的代码：oj测试通过 Runtime: 63 ms

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be inserted
    # @return integer
    def searchInsert(self, A, target):
        # zero length case
        if len(A) == 0 :
            return 0
        # binary search
        start = 0
        end = len(A)-1
        while start <= end :
            if start == end:
                if start == 0:
                    return [0,1][A[0]<target]
                if A[start] == target:
                    return start
                else:
                    return [start,start+1][A[start]<target]
            if start+1 == end:
                if A[start] >= target:
                    return start
                elif A[start] < target and A[end] >= target:
                    return end
                else:
                    return end+1
            mid = (start+end)/2
            if A[mid] == target:
                return mid
            elif A[mid] > target:
                end = mid - 1
            else:
                start = mid + 1

思路跟非递归差不太多。

个人感觉判断stop case的代码虽然逻辑上比较清晰（剩一个元素或者两个元素或者直接找到了target），但是并不是很简洁。后续再不断改进。