数据结构之二分法查找（左右边界，详细整理）

原创已于 2022-11-28 12:12:55 修改 · 2.3k 阅读

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本文详细介绍了二分法在数据结构中的应用，包括二分法查找、求左边界和求右边界的功能实现，分别给出了Python和C++的代码示例。通过这些例子，读者可以更好地理解和掌握二分法的运用。

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二分法作为数据结构的基础知识，经常作为使用，特意作为一次总结

强调：二分法有两种形式，主要差异在于right的限制，本博客主要以其中一种方式展开，这种方式便于记忆和理解

二分法查找

python

def binarySearch(nums: List[int], target[int]) -> int:
	left = 0, right = len(nums)-1
	while left <= right:
		mid = left + (right - left) // 2
		if nums[mid] > target: right = mid - 1
		elif nums[mid] < target: left = mid + 1
		else: return mid
	return -1

c++

int binarySearch(vector<int>& nums, int target) {
	int left = 0, right = nums.size()-1;
	while (left <= right) {
		int mid = left + (right - left) / 2;
		if (nums[mid] > target) right = mid - 1;
		else if (nums[mid] < target) left = mid + 1;
		else return mid;
	}
	return -1;
}

二分法求左边界

python

def leftBound(nums: List[int], target: int) -> int:
	left = 0
	right = len(nums)-1
	while left <= right:
		mid = left + (right - left) // 2
		if nums[mid] > target: right = mid - 1
		elif nums[mid] < target: left = mid + 1
		else: right = mid - 1
	if left >= len(nums) or nums[left] != target: return -1
	return left

c++

int leftBound(vector<int>& nums, int target) {
	int left = 0, right = nums.size()-1;
	while (left <= right) {
		int mid = left + (right - left) / 2;
		if (nums[mid] > target) right = mid - 1;
		else if (nums[mid] < target) left = mid + 1;
		else right = mid - 1;
	}
	if (left >= nums.size() || nums[left] != target) return -1;
	return left;
}

二分法求右边界

python

def rightBound(nums: List[int], target: int) -> int:
	left = 0
	right = len(nums)-1
	while left <= right:
		mid = left + (right - left) // 2
		if (nums[mid] > target) right = mid - 1
		elif (nums[mid] < target) left = mid + 1
		else: left = mid + 1
	if right < 0 or nums[right] != target: return -1
	return right

c++

int rightBound(vector<int>& nums, int target) {
	int left = 0, right = nums.size()-1;
	while (left <= right) {
		int mid = left + (right - left) / 2;
		if (nums[mid] > target) right = mid - 1;
		else if (nums[mid] < target) left = mid + 1;
		else left = mid + 1;
	}
	if (right < 0 || nums[right] != target) return -1;
	return right;
}

案例

二分法求方程的根

# 二分法求解y=sqrt(x)
# x = 1.0, y = 1.0; x = 3, y = 1.732; x = 0.01, y = 0.1

def solution(x):
    if x > 1:
        left, right = 0, x
    else:
        left, right = 0, 1

    while left <= right:
        mid = left + (right - left) / 2
        if mid * mid == x or abs(x-mid) < 10e-5:
            return mid
        elif mid * mid > x:
            right = mid - 0.001
        else:
            left = mid + 0.001
    return right



print(solution(0.01))