LeetCode之Remove Element
【摘要】 1、题目
Given an array and a value, 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 in place with c...
1、题目
Given an array and a value, 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 in place with constant memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
Example:
Given input array nums = [3,2,2,3], val = 3
Your function should return length = 2, with the first two elements of nums being 2.
2、代码实现
java:
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public class Solution {
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public int removeElement(int[] nums, int val) {
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if (nums == null || nums.length == 0)
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return 0;
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int count = 0;
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for (int i = 0; i < nums.length; ++i) {
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if (nums[i] != val) {
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nums[count++] = nums[i];
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}
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}
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return count;
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}
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}
C++:
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class Solution {
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public:
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int removeElement(vector<int>& nums, int val) {
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int count = 0;
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for (int i = 0; i < nums.size(); ++i) {
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if (nums[i] != val) {
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nums[count++] = nums[i];
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}
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}
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return count;
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}
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};
3、总结
其实非常简单，把不相等的数据从头到尾放进数组就行
文章来源: chenyu.blog.csdn.net，作者：chen.yu，版权归原作者所有，如需转载，请联系作者。
原文链接：chenyu.blog.csdn.net/article/details/77132335
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