Given a target number and an integer array A sorted in ascending order, find the index i in A such that A[i] is closest to the given target.
Return -1 if there is no element in the array.
分析
使用binary Search 找到可以插入 target 的 position, 例如是 i, 那么，从i（包括i）到后面，都是大于等于target的数字
1 如果 i == 0, return i，因为A[i]一定是最接近 target的数字，后面的数字都大于A[i]
2 如果 i > =，那么需要考虑A[i] 和 A[i - 1] 哪个更接近 target，就返回哪一个
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public class Solution {
/**
* @param A an integer array sorted in ascending order
* @param target an integer
* @return an integer
*/
public int closestNumber(int[] A, int target) {
// Write your code here
if(A == null || A.length == 0)
return -1;
int left = 0, right = A.length - 1, mid;
while(left < right){
mid = left + (right - left) / 2;
if(A[mid] < target){
left = mid + 1;
}
else{
right = mid;
}
}
// when left == right, this is the first position that target can be insert
if(right > 0 && (A[right] - target) > (target - A[right - 1]))
return right - 1;
else
return right;
}
} |