# binary search solution

• ``````public int findPairs(int[] nums, int k) {
Arrays.sort(nums);//first sort the array
int count = 0;
for (int i = 0; i < nums.length - 1; i++) {
/**
* since the array is sorted, it's ascending, and skip the number that has been searched
*/
if (i != 0 && nums[i] == nums[i - 1])continue;
/**
* since we are to find nums[i] - nums[j] == k, so just binary search the array from i + 1 for
* target nums[i] + k,to reason why there is no need to search from 0 to nums.length - 1 is
* that it must be searched before,if there exist j, that nums[j] =  nums[i] - k, so it must
* be calculated when i == j above,num[i] = nums[j] +k
*/
count += bianrySearch(nums, i + 1,nums[i] + k) ? 1 : 0;
}
return count;
}
private boolean bianrySearch(int[]nums, int low, int target) {
int high = nums.length - 1;
while (low <= high) {
int med = (low + high) >> 1;
if (nums[med] > target) high = med - 1;
else if (nums[med] < target) low = med + 1;
else return true;
}
return false;
}
``````

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