leetcode-8

Climbing Stairs

Just Fibonacci. Dynamic Programming's space complexity can be optimized by only store previous two value.

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Sort Colors

1. traverse one pass, counting all elements then rearrange.
2. 3 pointers

    void swap(int &a, int &b){
        int temp = a;
        a=b;
        b=temp;
    }
public:
    void sortColors(vector& nums) {
        int low = 0;
        int high = nums.size()-1;
        int mid = 0;
        while(mid<=high){
            if(nums[mid]==0){
                swap(nums[mid],nums[low]);
                low++;
                mid=low;
            }
            else if(nums[mid]==2){
                swap(nums[mid], nums[high]);
                high--;
            }
            else
            mid++;
        }
    }

When swap in place, one thing need to pay attention is that the swapped value will change to an unknown value, so mid cannot move, need to judge again.

3. Still 3 pointers

void sortColors(int A[], int n) {
    int n0 = -1, n1 = -1, n2 = -1;
    for (int i = 0; i < n; ++i) {
        if (A[i] == 0) 
        {
            A[++n2] = 2; A[++n1] = 1; A[++n0] = 0;
        }
        else if (A[i] == 1) 
        {
            A[++n2] = 2; A[++n1] = 1;
        }
        else if (A[i] == 2) 
        {
            A[++n2] = 2;
        }
    }
}

All pointers begin at 0 and then add up.

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