LeetCode完全平方数问题：0-9999整数能否安全用float替代int？
2026-6-19
完全平方数问题中float替代int的性能与兼容性疑问
在LeetCode的「完全平方数」问题求解过程中，我尝试用float类型替代int类型后获得了性能提升。当整数取值范围保证为0≤数值<10000时，float与int版本的代码均通过了所有测试，但我不确定是否会因为CPU、编译器差异或其他因素（如执行时间波动）引发潜在问题。
示例比较代码
if(n == i*i + j*j * 2) result3++; if(n == i*i + k*k) result2++;
实现代码
#include<iostream> #include<math.h> class Solution { public: static constexpr int simd =8; using FAST_TYPE = short; using MASK_TYPE = short; const int numSquares(const int n) const noexcept { if(n==2 || n==8) return 2; if(n==3 || n==6 || n==11) return 3; if((int)std::sqrt(n)*(int)std::sqrt(n) == n) return 1; FAST_TYPE found2 = 0; FAST_TYPE found3 = 0; FAST_TYPE found32 = 0; FAST_TYPE found33 = 0; FAST_TYPE found34 = 0; alignas(64) FAST_TYPE zeroSimd[simd]; alignas(64) FAST_TYPE oneSimd[simd]; alignas(64) FAST_TYPE found3Simd[simd]; alignas(64) FAST_TYPE found3Simd2[simd]; alignas(64) FAST_TYPE found3Simd3[simd]; alignas(64) FAST_TYPE found3Simd4[simd]; alignas(64) FAST_TYPE mSimd[simd]; alignas(64) FAST_TYPE kSimd[simd]; alignas(64) FAST_TYPE k0Simd[simd]; alignas(64) FAST_TYPE nSimd[simd]; alignas(64) FAST_TYPE twoSimd[simd]; alignas(64) FAST_TYPE threeSimd[simd]; alignas(64) FAST_TYPE iSimd[simd]; alignas(64) FAST_TYPE jSimd[simd]; alignas(64) FAST_TYPE ijSimd[simd]; alignas(64) FAST_TYPE j2Simd[simd]; alignas(64) FAST_TYPE i2Simd[simd]; alignas(64) MASK_TYPE mask1Simd[simd]; alignas(64) MASK_TYPE mask2Simd[simd]; alignas(64) MASK_TYPE mask3Simd[simd]; alignas(64) MASK_TYPE mask4Simd[simd]; alignas(64) FAST_TYPE sum1Simd[simd]; alignas(64) FAST_TYPE sum2Simd[simd]; alignas(64) FAST_TYPE sum3Simd[simd]; alignas(64) FAST_TYPE mulSimd[simd]; for(int i=0;i<simd;i++) { zeroSimd[i]=0; oneSimd[i]=1; found3Simd[i]=0; found3Simd2[i]=0; found3Simd3[i]=0; found3Simd4[i]=0; mSimd[i]=i; nSimd[i]=n; twoSimd[i]=2; threeSimd[i]=2; } for(int i=1+std::sqrt(n);i>=1;i--) { const FAST_TYPE i2 = i*i; const FAST_TYPE i22 = 2*i*i; const FAST_TYPE i23 = 3*i*i; #pragma GCC ivdep for(int m=0;m<simd;m++) iSimd[m]=i2; #pragma GCC ivdep for(int m=0;m<simd;m++) i2Simd[m]=i22; found2 += (i22 == n); found3+=(i23 == n); for(int j=i-1;j>=1;j--) { const FAST_TYPE j2 = j*j; const FAST_TYPE j22 = 2*j*j; const FAST_TYPE j23 = 3*j*j; #pragma GCC ivdep for(int m=0;m<simd;m++) jSimd[m]=j2; #pragma GCC ivdep for(int m=0;m<simd;m++) j2Simd[m]=j22; #pragma GCC ivdep for(int m=0;m<simd;m++) ijSimd[m]=i2+j2; found2+=(i2 + j2 == n); found3+=(i2 + j22 == n)+(i22 + j2 == n)+(j23 == n); const int k32 = j-1 - ((j-1)%simd); #pragma GCC unroll 2 for(int k0=1;k0<=k32;k0+=simd) { #pragma GCC ivdep for(int m=0;m<simd;m++) k0Simd[m]=k0; #pragma GCC ivdep for(int m=0;m<simd;m++) kSimd[m] = k0Simd[m]+mSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) kSimd[m] = kSimd[m]*kSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) sum1Simd[m]=ijSimd[m] + kSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) mask1Simd[m]=sum1Simd[m] == nSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) found3Simd[m]=mask1Simd[m]?oneSimd[m]:found3Simd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) sum2Simd[m]=i2Simd[m] + kSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) mask2Simd[m]=(sum2Simd[m]==nSimd[m]); #pragma GCC ivdep for(int m=0;m<simd;m++) found3Simd2[m]=mask2Simd[m]?oneSimd[m]:found3Simd2[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) sum3Simd[m]=j2Simd[m] + kSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) mask3Simd[m]=(sum3Simd[m]==nSimd[m]); #pragma GCC ivdep for(int m=0;m<simd;m++) found3Simd3[m]=mask3Simd[m]?oneSimd[m]:found3Simd3[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) mulSimd[m]=threeSimd[m]*kSimd[m]; #pragma GCC ivdep for(int m=0;m<simd;m++) mask4Simd[m]=(mulSimd[m]==nSimd[m]); #pragma GCC ivdep for(int m=0;m<simd;m++) found3Simd4[m]=mask4Simd[m]?oneSimd[m]:found3Simd4[m]; } for(int k=k32;k<=j-1;k++) { const FAST_TYPE k2 = k*k; found3+=(i2 + j2 + k2 ==n); found32+=(i22 + k2 ==n); found33+=(j22 + k2 ==n); found34+=(3*k2 ==n); } } } for(int i=0;i<simd;i++) { found3+=found3Simd[i]; found32+=found3Simd2[i]; found33+=found3Simd3[i]; found34+=found3Simd4[i]; } found3 += found32 + found33 + found34; if(found2) return 2; if(found3) return 3; return 4; } }; int main() { Solution s; for(int i=10;i<20;i++) { std::cout<<i<<" is equal to sum of "<<s.numSquares(i)<< " perfect squares"<<std::endl; } }
输出结果
10 is equal to sum of 2 perfect squares 11 is equal to sum of 3 perfect squares 12 is equal to sum of 3 perfect squares 13 is equal to sum of 2 perfect squares 14 is equal to sum of 3 perfect squares 15 is equal to sum of 4 perfect squares 16 is equal to sum of 1 perfect squares 17 is equal to sum of 2 perfect squares 18 is equal to sum of 2 perfect squares 19 is equal to sum of 3 perfect squares
内容的提问来源于stack exchange，提问作者huseyin tugrul buyukisik
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