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
tag : binary search .
left < right - 1 目的是最终不会出现 left > right的情形。 因此，结束时可能最终指针停留在两个元素上。
public class Solution {
public int searchInsert(int[] nums, int target) {
if(nums == null ||nums.length == 0) {
return 0;
}
int left = 0;
int right = nums.length - 1;
while (left < right - 1) {
int mid = left + (right - left) / 2;
if(target == nums[mid]) {
return mid;
}
if(target > nums[mid]) {
left = mid;
} else {
right = mid;
}
}
if(nums[left] >= target) {
return left;
}
if(target > nums[right]) {
return right + 1;
}
return right;
}
}
leetcode : Search Insert Position
原文：http://www.cnblogs.com/superzhaochao/p/6407246.html