思路:二分搜索,对于当前元素A[mid],如果A[mid] > target,则应该从left 到 mid-1进行搜索,但是如果A[mid-1] < target 则插入位置就是mid这里,返回mid值,注意向左搜到mid==0时,还没有确定好位置,就应该返回0,并入到一起就是return mid,右边情况的分析类似。
code:
class Solution {
public:
int searchInsert(int A[], int n, int target) {
int left = 0, right = n-1;
while(left <= right){
int mid = (left + right)>>1;
if(A[mid] == target)return mid;
else if(A[mid] > target){
if(mid == 0 || A[mid-1] < target)return mid;
right = mid - 1;
}
else{
if(mid == n-1 || A[mid+1] > target)return mid+1;
left = mid + 1;
}
}
}
};