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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 = 3Your function should return length = 2, with the first two elements of nums being 2. |
题意:给一个数组,再给你一个数val,从数组中把val删掉并返回剩余数组的长度
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int removeElement(int* nums, int numsSize, int val) {
int i=0;
int j=0;
//这个就跟那个26题去重一个思想了
for(j=0;j<numsSize;j++){
if(nums[j]!=val){
nums[i++]=nums[j];
}
}
return i;
} |
PS:这个跟之前的26题一样,双指针,,,,,
本文转自 努力的C 51CTO博客,原文链接:http://blog.51cto.com/fulin0532/1868692
本文介绍了一个简单的算法,用于从数组中删除指定值的所有实例,并返回处理后数组的新长度。该算法采用双指针技巧,无需额外空间,在原地修改数组。
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