35. 搜索插入位置

Problem: 35. 搜索插入位置

解析

二分查找的简单应用

二分查找 递归法

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//! log(n),log(n)->二分查找log(n)次,函数递归栈log(n)大小
class Solution {
public:
int searchInsert(vector<int> &nums, int target) {
return binarySearch(nums, 0, nums.size() - 1, target);
}
int binarySearch(vector<int> nums, int left, int right, int target) {
if (left > right) {
return left;
}
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] > target) {
return binarySearch(nums, left, mid - 1, target);
} else {
return binarySearch(nums, mid + 1, right, target);
}
}
};

二分查找 迭代法

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//! log(n),1->二分查找log(n)次,常数空间
class Solution {
public:
int searchInsert(vector<int> &nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid - 1;
} else {
return mid;
}
}
return left;
}
};