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Copy pathBinary_search_range.cpp
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Binary_search_range.cpp
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// 一道求数组中目标数上确界和下确界的题目
// input: nums = [5,7,7,8,8,10], target = 8
// output: [3, 4]
#include <vector>
class Solution {
public:
vector<int> serarchRange(vector<int>& nums, int target) {
// 找目标数的下确界
int index1 = findTarget(nums, target);
// 小tricky,找目标数+1的下确界然后再减一,就是目标数的上确界
int index2 = findTarget(nums, target + 1) - 1;
// 错误处理,如果数组中没有目标数,返回{-1,-1}
if (index1 < nums.size() && nums[index1] == target) {
return {index1, index2}
}
else{
return {-1, -1}
}
}
// 找下确界,二分法,注意搜索范围,循环终止条件
int findTarget(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1;
// 这里就定义了搜索区间是[left, right],循环终止条件是left > right,也就是说搜索区间变为[right + 1, right],
// 这个区间确实不能搜索了,如果是left < right,循环终止条件变为left >= right,
// 搜索区间中的[right, right]就没被处理到
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
right = mid - 1; //找下确界,不断往左逼近
}
else if (nums[mid] < target) {
left = mid + 1;
}
else if (nums[mid] > target) {
right = mid - 1;
}
}
return left;
}
};