Leetcode之Remove Element
阿新 • • 發佈:2018-12-28
題目:
Given an array nums and a value val, remove all instances of that value in-place and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
Example 1:
Given nums = [3,2,2,3], val = 3, Your function should return length = 2, with the first two elements of nums being 2. It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2, Your function should return length =5
, with the first five elements ofnums
containing0
,1
,3
,0
, and 4. Note that the order of those five elements can be arbitrary. It doesn't matter what values are set beyond the returned length.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy) int len = removeElement(nums, val); // any modification to nums in your function would be known by the caller. // using the length returned by your function, it prints the first len elements. for (int i = 0; i < len; i++) { print(nums[i]); }
程式碼:
class Solution {
public:
int removeElement(vector<int>& nums, int val) {
int len = nums.size();
if (len <= 0)return 0;
int current=-1, ok = 0;
for (int i = 0; i < len; i++) {
if ((nums[i] == val) && (ok==0)) {
current = i;
ok = 1;
}
if ((nums[i] != val) && (current != -1)) {
int temp = nums[i];
nums[i] = nums[current];
nums[current] = temp;
current++;
}
}
int i;
for (i = 0; i < len; i++) {
if (nums[i] == val)break;
}
return i;
}
};
注意:
1、要考慮待刪的值是第一個元素的情況。
2、不要粗心,看清楚條件內的語句和表達的意思是否一致。(ok等於0才進去)