【leetcode】13. Roman to Integer(C)

题目链接

提交代码:

int romanToInt(char* s) {
  int i,p = 0,numslen=0;
	int sum = 0;
	int* nums = (int*)malloc(30 * sizeof(int));

	while (s[p] != '\0')
	{
		if (s[p] == 'M')
			nums[p] = 1000;
		else if (s[p] == 'D')
			nums[p] = 500;
		else if (s[p] == 'C')
			nums[p] = 100;
		else if (s[p] == 'L')
			nums[p] = 50;
		else if (s[p] == 'X')
			nums[p] = 10;
		else if (s[p] == 'V')
			nums[p] = 5;
		else if (s[p] == 'I')
			nums[p] = 1;
		p++;
	}
	numslen=p;

	for (i = 0; i < numslen; i++)
	{
		//如果前一个数大于当前数,说明当前数是要减的
		if (nums[i + 1] > nums[i])
			sum -= nums[i];
		else
			sum += nums[i];
	}
	
	return sum;  
}

运行结果:

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转载自blog.csdn.net/AXIMI/article/details/83141691