题目描述
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
解题思路
方法一
代码冗余，并且测试用例有一种情况没通过
解答失败:
测试用例:[5,7,7,8,8,10]
6
测试结果:1
期望结果:0
--------------> 有BUG
public int search_1(int[] nums, int target) {
int start = 0, end = nums.length - 1;
int count_start = 0,count_end = 0;
int res;
for (int i = 0; i < nums.length - 1; i++) {
int mid = (end - start) /2 + start;
if (nums[mid] == target){
count_start = mid;
count_end = mid;
if (nums[count_start] == target){
count_start --;
}else if (nums[count_end] == target){
count_end ++;
}
}else if (nums[mid] < target){
start = mid + 1;
}else if (nums[mid] > target){
end = mid - 1;
}
}
res = count_end - count_start + 1;
return res;
}
方法二：
一个数的出现次数 = 这个数的开始下标 -下一个数的开始下标
由于数组 nums 中元素都为整数，因此可以分别二分查找 target 和 target +1 的右边界，将两结果相减并返回即可。
public int search_2(int[] nums,int target){
if (nums == null){
return 0;
}
return binarySearch(nums,target + 1) - binarySearch(nums,target);
}
private int binarySearch(int[] nums, int target) {
int leftIndex = 0;
int rightIndex = nums.length - 1;
while (leftIndex <= rightIndex){
int midIndex = leftIndex + (rightIndex - leftIndex) /2;
if (nums[midIndex] < target){
leftIndex = midIndex + 1;
}else {
rightIndex = midIndex - 1;
}
}
return leftIndex;
}
方法三：
初始化：low = 0, high = nums.length - 1.
当low≤high时递归二分查找：
mid = (low + high) / 2
若nums[mid] == target，[low, mid - 1]和[mid + 1, high]都有可能还有target，因为存在重复元素，继续递归查找。
若nums[mid] < target，说明target只可能出现在区间[mid + 1, high]，继续递归查找。
若nums[mid] > target，说明target只可能出现在区间[low, mid - 1]，继续递归查找。
算法结束，返回计数器的值
int cnt = 0; //计数器
public int search(int[] nums,int target){
helper(nums,target,0,nums.length - 1);
return cnt;
}
private void helper(int[] nums, int target, int low, int high) {
if (low <= high){
int mid = low + (high - low) /2;
if (nums[mid] == target){
cnt++; //计数一次
helper(nums,target,low,mid - 1);
helper(nums,target,mid + 1,high);
}else if (nums[mid] > target){
helper(nums,target,low,mid - 1);
}else{
helper(nums,target,mid + 1,high);
}
}
}