Description
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.
Solution
移除数组中和指定val相等的元素,返回剩余元素个数k,和26. Remove Duplicates from Sorted Array类似,保证前k个元素ok就行,后面n-k位置放什么不用管
int removeElement(vector<int>& nums, int val) {
if (nums.empty()) {
return 0;
}
int index = 0;
for (int i = 0; i < nums.size(); ++i) {
//如果找到OK的元素,填上坑,言外之意是对于需要删除的元素的坑,无所谓谁去填
if (nums[i] != val) {
nums[index++] = nums[i];
}
}
return index;
}
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