{"id":8785,"date":"2021-11-26T13:48:08","date_gmt":"2021-11-26T21:48:08","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8785"},"modified":"2021-11-26T13:52:26","modified_gmt":"2021-11-26T21:52:26","slug":"leetcode-29-divide-two-integers","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/bit\/leetcode-29-divide-two-integers\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 29. Divide Two Integers"},"content":{"rendered":"\n

Given two integers dividend<\/code> and divisor<\/code>, divide two integers without<\/strong> using multiplication, division, and mod operator.<\/p>\n\n\n\n

The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345<\/code> would be truncated to 8<\/code>, and -2.7335<\/code> would be truncated to -2<\/code>.<\/p>\n\n\n\n

Return the quotient<\/strong> after dividing <\/em>dividend<\/code> by <\/em>divisor<\/code>.<\/p>\n\n\n\n

Note: <\/strong>Assume we are dealing with an environment that could only store integers within the 32-bit<\/strong> signed integer range: [\u2212231<\/sup>, 231<\/sup> \u2212 1]<\/code>. For this problem, if the quotient is strictly greater than<\/strong> 231<\/sup> - 1<\/code>, then return 231<\/sup> - 1<\/code>, and if the quotient is strictly less than<\/strong> -231<\/sup><\/code>, then return -231<\/sup><\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> dividend = 10, divisor = 3\nOutput:<\/strong> 3\nExplanation:<\/strong> 10\/3 = 3.33333.. which is truncated to 3.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> dividend = 7, divisor = -3\nOutput:<\/strong> -2\nExplanation:<\/strong> 7\/-3 = -2.33333.. which is truncated to -2.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> dividend = 0, divisor = 1\nOutput:<\/strong> 0\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> dividend = 1, divisor = 1\nOutput:<\/strong> 1\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • -231<\/sup> <= dividend, divisor <= 231<\/sup> - 1<\/code><\/li>
  • divisor != 0<\/code><\/li><\/ul>\n\n\n\n
    Solution: Binary \/ Bit Operation<\/strong><\/h2>\n\n\n\n
    The answer can be represented in binary format.<\/p>\n\n\n\n
    a \/ b = c
    a >= b * \u00a0\u03a3(di<\/sub>*2i<\/sup>)
    c = \u03a3(di<\/sub>*2i<\/sup>)<\/p>\n\n\n\n
    e.g. 1: 10 \/ 3
    => 10 >= 3*(21<\/sup> + 20<\/sup>) = 3 * (2 + 1) = 9
    ans = 21<\/sup> + 20<\/sup> = 3

    e.g. 2: 100 \/ 7
    => 100 >= 7*(23<\/sup> + 22<\/sup>+21<\/sup>) = 7 * (8 + 4 + 2) = 7 * 14 = 98
    ans = 23<\/sup>+22<\/sup>+21<\/sup>=8+4+2=14<\/p>\n\n\n\n
    Time complexity: O(32)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int divide(int A, int B) {\n    if (B == INT_MIN)\n      return A == INT_MIN ? 1 : 0;\n    if (A == INT_MIN) {\n      if (B == -1) return INT_MAX;\n      return B > 0 ? divide(A + B, B) - 1 : divide(A - B, B) + 1;\n    }    \n    int a = abs(A);\n    int b = abs(B);\n    int ans = 0;\n    for (int x = 31; x >= 0; --x)\n      if (a >> x >= b)\n        ans += 1 << x, a -= b << x;\n    return (A > 0) == (B > 0) ? ans : -ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator. The integer division should truncate toward zero, which means losing its fractional part. For…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126],"tags":[39,16,177],"class_list":["post-8785","post","type-post","status-publish","format-standard","hentry","category-bit","tag-binary","tag-bit","tag-medium","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8785","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=8785"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8785\/revisions"}],"predecessor-version":[{"id":8788,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8785\/revisions\/8788"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8785"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8785"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8785"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}