Lintcode 14 first position of target
2018-03-18 09:33
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题目:For a given sorted array (ascending order) and a
解析:这个题考的是二分查找,但是有一点不同的是要求返回符合的第一个数在数组中的位置
代码如下:public class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts from 0.
*/
public int binarySearch(int[] nums, int target) {
// write your code here
int mid = 0;
int start = 0;
int end = nums.length-1;
while(start<end)
{
mid = (end + start)/2;
if(nums[mid]>target)
end = mid - 1;
if(nums[mid]<target)
start = mid + 1;
else if(nums[mid]==target)
end = mid;
}
if(nums[end]==target)
return end;
return -1;
}
}
后期需要完善的是:判断数组为空或者是长度为0的情况,可以节省时间复杂度
targetnumber, find the first index of this number in
O(log n)time complexity.If the target number does not exist in the array, return
-1.ExampleIf the array is
[1, 2, 3, 3, 4, 5, 10], for given target
3, return
2.
解析:这个题考的是二分查找,但是有一点不同的是要求返回符合的第一个数在数组中的位置
代码如下:public class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts from 0.
*/
public int binarySearch(int[] nums, int target) {
// write your code here
int mid = 0;
int start = 0;
int end = nums.length-1;
while(start<end)
{
mid = (end + start)/2;
if(nums[mid]>target)
end = mid - 1;
if(nums[mid]<target)
start = mid + 1;
else if(nums[mid]==target)
end = mid;
}
if(nums[end]==target)
return end;
return -1;
}
}
后期需要完善的是:判断数组为空或者是长度为0的情况,可以节省时间复杂度
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