34. Search for a Range

建议先看这里：二分查找。

Problem

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution

one

WTF 折腾了大半早上

本来是直接定义low/high为decltype(nums.size())，但是，当出现-1时候，出问题啊，decltype(nums.size())是unsigned long int…

这下对二分查找算是整的清楚了，以后边界定义就按照这个来，二分查找，就不用!=了。

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ret(2, -1);        int n = nums.size();        int low = 0, high = n-1, mid;        while (low <= high) {            mid = low + (high - low) / 2;            if (nums[mid] == target) {                low = mid; //找到后，往前/后继续查找范围                high = mid;                while (low != -1 && nums[low] == target) {                    --low;                }                ret[0] = low+1;                while (high != n && nums[high] == target) {                    ++high;                }                ret[1] = high-1;                return ret;            }            else if (nums[mid] < target) {                low = mid + 1;            }            else {                high = mid - 1;            }        }        return ret;    }};

runtime: 9ms

时间貌似挺快的

two

date: 2016.9.16

思路请看关于二分

class Solution {public:    int upper_bound(vector<int> &nums, int low, int high, int target) {        int mid;        while (low < high) {            mid = low + (high-low)/2;            if (nums[mid] <= target) {                 low = mid+1;            }            else {                high = mid;            }        }        return low;    }    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> v(2, -1);        int n = (int)nums.size();        if (n == 1 && nums[0] != target) {            return v;        }        if (n == 1 && nums[0] == target) {            v[0] = 0;            v[1] = 0;            return v;        }        if (nums[n-1] < target) {            return v;        }        if (nums[0] > target) {            return v;        }        int low = 0, high = n;        high = upper_bound(nums, low, high, target);        if (nums[--high] != target) {            return v;        }        v[1] = high;        while (high >= 0 && nums[high] == target) {            --high;        }        if (high >= 0) {            v[0] = ++high;            return v;        }        v[0] = 0;        return v;    }};

runtime: 9ms

关于二分

date: 2016.9.16

二分查找，花了不少时间，重新写一份关于二分查找的吧，算是给二分一个结局，地址：二分查找。