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看到网上一些资料的案例不全,所以自己开个来复习。 O(1)<O(log2n)<O(n)<O(nlog2n)<O(n^2)<O(n^3)<…<O(2^n)<O(n!)
O(1) int a = 1, b = 3, c = 3; O(log2n) int i = 1; while(i<=n) { i*=2 } O(n) int sum = 0; for(int i = 0; i < n; i++) { sum+=i; } O(nlog2n) for(int m=1;m<n;m++) { i=1; while(i<n) { i=i*2; } } O(n^2) int num1, num2; for(int i=0; i<n; i++) { num1 += 1; for(int j=1; j<=n; j++) { num2 += j; } } O(n^3) int num1, num2, num3; for(int i=0; i<n; i++) { num1 += 1; for(int j=1; j<=n; j++) { num2 += j; for(int k=1; k<=n; k++) { num3 += k; } } } O(2^n) 2^n = 2*2*2*........2* // 有n个2 O(n!) n! = 1*2*3*4*5......*n; 相关文章: https://blog.csdn.net/u012925946/article/details/84030160 https://blog.csdn.net/qq_41672557/article/details/101699850 https://blog.csdn.net/qq_40513633/article/details/107705801
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