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June 23, 2016 18:49
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/** | |
* Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. | |
* You may assume no duplicates in the array. | |
* Here are few examples. | |
* [1,3,5,6], 5 → 2 | |
* [1,3,5,6], 2 → 1 | |
* [1,3,5,6], 7 → 4 | |
* [1,3,5,6], 0 → 0 | |
*/ | |
public class Solution{ | |
public int searchInsert(int[] A, int target) { | |
if(A == null || A.length == 0) { | |
return 0; | |
} | |
int start = 0; | |
int end = A.length - 1; | |
int mid = 0; | |
while(start <= end){ | |
mid = start + (end - start) / 2; | |
if(A[mid] == target) { | |
return mid; | |
} else if(A[mid] < target){ | |
// search from the right part | |
start = mid + 1; | |
}else { | |
// search from the left part | |
end = mid - 1; | |
} | |
} | |
if(A[mid] < target) { | |
return mid + 1; | |
} | |
return mid; | |
} | |
} | |
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