LeetCode (35)Search Insert Position
2017-06-22 13:39
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(35)Search Insert Position
题目:在一个排好序的数组里,找目标数字位置,如果找到目标数字就返回下标,找不到数字就返回它应该插在的位置下标。例子:
序列: 目标数字: 返回值: [1, 3, 5, 6] 5 2 [1, 3, 5, 6] 2 1 [1, 3, 5, 6] 7 4 [1, 3, 5, 6] 0 0
根据题目就使用二分法每次对比中位数就可以了,因为递归做到最后肯定顶多只剩两个数字,这两个数字要么一个比目标大,一个比目标小,要么两个都比目标小或者大,此时只要考虑mid和mid+1两个数字的关系,由于还会存在目标数字的大小是可能在数组两端的情况,所以还要考虑mid的值是否是0或者是mid+1与nums.size()-1相同,且mid+1位置的数字小于当前目标数字,那么就要插在最右侧。
下面是代码:
class Solution { public: int searchint(vector<int> nums, int lef, int rig, int target){ int mid = (lef + rig) / 2; if(nums[mid] == target){ return mid; else if(nums[mid] > target){ if(mid == 0){ return 0; } return searchint(nums, lef, mid, target); } else{ if(nums[mid+1] == target){ return mid + 1; } else if(nums[mid+1] > target){ return mid+1; } else{ if(mid + 1 == nums.size()-1){ return mid + 2; } return searchint(nums, mid, rig, target); } } } int searchInsert(vector<int>& nums, int target) { if(nums.size() == 1){ if(nums[0] >= target){ return 0; } else{ return 1; } } return searchint(nums, 0, nums.size()-1, target); } };
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