leetCode- Remove Element

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 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:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

我的版本(适合删除元素很多的情况):

class Solution {
public int removeElement(int[] nums, int val) {
int count=0;

for(int i=0;i<nums.length;i++){
if(nums[i]!=val){
nums[count++] = nums[i];
}
}
return count;
}
}

其他版本(适合删除元素很少的情况):

public int removeElement(int[] nums, int val) {
    int i = 0;
    int n = nums.length;
    while (i < n) {
        if (nums[i] == val) {
            nums[i] = nums[n - 1];
            // reduce array size by one
            n--;
        } else {
            i++;
        }
    }
    return n;
}

leetCode- Remove Element

原文：http://www.cnblogs.com/kevincong/p/7802602.html