leetcode: Remove Element
来源:互联网 发布:买mac还是ipad 编辑:程序博客网 时间:2024/05/29 02:31
Given an array and a value, remove all instances of that value in place and return the new length.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
解法一:
class Solution {public: int removeElement(int A[], int n, int elem) { int nl, i; if (n == 0) { return n; } nl = n; for (i = 0; ;) { if (A[i] == elem) { while (A[nl-1] == elem) { //nl--; if (i == nl - 1) { return nl - 1; } nl--; } A[i] = A[nl-1]; nl--; } i++; if (i >= nl) { break; } } return nl; }};
解法二(很巧妙的解法)
class Solution {public: int removeElement(int A[], int n, int elem) { int len, i, pos; if (n == 0) { return n; } len = n; for (i = 0, pos = 0; i < n; i++) {/* just like double arrays */ if (A[i] == elem) { len--; } else { A[pos++] = A[i]; } } return len; }};
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