面试题53 - I. 在排序数组中查找数字 I
统计一个数字在排序数组中出现的次数。
示例 1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: 2
示例 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: 0
限制:
0 <= 数组长度 <= 50000
注意:本题与主站 34 题相同(仅返回值不同):https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/
测试用例:
[]
1
[1]
1
[1]
0
[5,7,7,8,8,10]
8
[5,7,7,8,8,10]
6
思路:二分法
class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size(), start = 0, end = 0;
int left = 0, right = n, mid = 0;
while (left < right) {
mid = left + (right - left) / 2;
if (nums[mid] == target) {
right = mid;
} else if (nums[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
if (left == n) return 0;
if (nums[left] == target) start = left;
else return 0;
left = 0, right = n, mid = 0;
while (left < right) {
mid = left + (right - left) / 2;
if (nums[mid] == target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
//if (left == 0) return 0;
if (nums[left-1] == target) end = left - 1;
//else return 0;
return end - start + 1;
}
};
/*12ms,15.7MB*/
比上面解法效果进一步优化:
class Solution {
public int result = 0;
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
while (left >= 0 && right <= nums.length - 1 && left <= right) {
int mid = (left + right) / 2;
if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid - 1;
} else {
result = 1;
int _left = mid - 1;
int _right = mid + 1;
while (_left >= left) {
if (nums[_left--] == target) result++;
else break;
}
while (_right <= right) {
if (nums[_right++] == target) result++;
else break;
}
break;
}
}
return result;
}
}
作者:le-ge-zhi-shang
链接:https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/solution/shuang-100ji-bai-er-fen-cha-zhao-fa-by-le-ge-zhi-s/
来源:力扣(LeetCode)
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