LEET code -- Remove Element

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.

之前leet code有類似的題目
一個pointer去找val 另外一個去找非val的  在交換

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int removeElement(int* nums, int numsSize, int val) { int Nonval=0,Isval=0; for(int i=0;i<numsSize;i++){ while(Isval<numsSize && nums[Isval]!=val)Isval++; //while(Nonval<numsSize && nums[Nonval]==val)Nonval++; Nonval=Isval; while(Nonval<numsSize && nums[Nonval]==val)Nonval++; if(Nonval>=numsSize)return Isval-1>=0? Isval:0; else{ int temp = nums[Nonval]; nums[Nonval]=nums[Isval]; nums[Isval]=temp; } Isval++; } }

更漂亮的寫法

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int removeElement(int* nums, int numsSize, int val) { int remove = 0; for(int i = 0; i < numsSize; ++i) { if(nums[i] == val) ++remove; else nums[i - remove] = nums[i]; } return numsSize - remove; }