LeetCode 288 周赛AC*2

最新推荐文章于 2025-05-22 00:33:56 发布

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本文记录了一位常胜选手在比赛中成功解答两道代码挑战的经历，涉及字符串处理与最小化表达式的优化技巧。

作为一个一直1AC的选手来说，能够A掉两题真的超级开心，可能是前两题比较简单吧，就直接上截图吧。

第一题代码

func largestInteger(num int) int {
	s := strconv.Itoa(num)
	n := len(s)
	odd, even := []int{}, []int{}
	vis := []int{}
	for i := 0; i < n; i++ {
		if s[i]%2 == 0 {
			odd = append(odd, -int(s[i]))
			vis = append(vis, 1)
		} else {
			even = append(even, -int(s[i]))
			vis = append(vis, 0)
		}
	}
	sort.Ints(odd)
	sort.Ints(even)
	res := ""
	i, j := 0, 0
	for k := 0; k < n; k++ {
		if vis[k] == 1{
			res += string(-odd[i])
			i++
		}else{
			res += string(-even[j])
			j++
		}
	}
	ans, _ := strconv.Atoi(res)
	return ans
}

第二题代码

func minimizeResult(expression string) string {
	nums := strings.Split(expression, "+")
	n1, n2 := nums[0], nums[1]
	nc1, nc2 := len(n1), len(n2)
	min := math.MaxInt32
	idxI, idxJ := 0, 0
	for i := 0; i < nc1; i++ {
		n11, n12 := 0, 0
		if i == 0 {
			n12, _ = strconv.Atoi(n1)
		} else {
			n11, _ = strconv.Atoi(n1[:i])
			n12, _ = strconv.Atoi(n1[i:])
		}
		for j := 1; j <= nc2; j++ {
			n21, n22 := 0, 0
			if j == nc2 {
				n21, _ = strconv.Atoi(n2)
			} else {
				n21, _ = strconv.Atoi(n2[:j])
				n22, _ = strconv.Atoi(n2[j:])
			}

			if n11 == 0 {
				n11 = 1
			}
			if n22 == 0 {
				n22 = 1
			}
			sum := n11 * (n12 + n21) * n22
			//fmt.Printf("%d*(%d+%d)*%d=%d\n", n11, n12, n21, n22, sum)
			if sum < min {
				min = sum
				idxI, idxJ = i, j+nc1
			}
		}
	}
	//fmt.Printf("idxi=:%d,idxj=%d\n", idxI, idxJ)
	//fmt.Printf("min=%d\n", min)
	return expression[:idxI] + "(" + expression[idxI:idxJ+1] + ")" + expression[idxJ+1:]
}