给定二进制字符串中最左侧设置位的位置，其中所有 1 都出现在末尾

给定一个长度为N的二进制字符串S ，使得所有 1 都出现在右侧。任务是返回从左侧找到的第一个设置位的索引，否则返回-1。

例子：

Input: s = 00011, N = 5
Output: 3
Explanation: The first set bit from the left side is at index 3.

Input: s = 0000, N = 4
Output: -1

方法：这个问题可以使用二分搜索来解决。

• 在 [1, N] 范围内应用二分查找来查找第一个设置位，如下所示：
• 将mid更新为 (l+r)/2
• 如果设置了 s[mid] 位，则将ans更新为 mid 并将r更新为mid-1
• 否则将l更新为 mid + 1
• 返回ans中最后存储的值。

下面是上述方法的实现：

C++

// C++ Program to find Position
// Of leftmost set bit
#include 
using namespace std;
 
// Function to find
// A bit is set or not
bool isSetBit(string& s, int i)
{
    return s[i] == '1';
}
 
// Function to find
// First set bit
int firstSetBit(string& s, int n)
{
    long l = 0, r = n, m, ans = -1;
 
    // Applying binary search
    while (l <= r) {
 
        m = (l + r) / 2;
        if (isSetBit(s, m)) {
 
            // store the current
            // state of m in ans
            ans = m;
            r = m - 1;
        }
        else {
            l = m + 1;
        }
    }
 
    // Returning the position
    // of first set bit
    return ans;
}
 
// Driver code
int main()
{
 
    string s = "111";
    int n = s.length();
    cout << firstSetBit(s, n);
 
    return 0;
}

Java

// Java program for the above approach
class GFG
{
 
  // Function to find
  // A bit is set or not
  static boolean isSetBit(String s, int i)
  {
    return s.charAt(i) == '1' ? true : false;
  }
 
  // Function to find
  // First set bit
  static int firstSetBit(String s, int n)
  {
    int l = 0, r = n, m, ans = -1;
 
    // Applying binary search
    while (l <= r) {
 
      m = (l + r) / 2;
      if (isSetBit(s, m)) {
 
        // store the current
        // state of m in ans
        ans = m;
        r = m - 1;
      }
      else {
        l = m + 1;
      }
    }
 
    // Returning the position
    // of first set bit
    return ans;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    String s = "111";
    int n = s.length();
    System.out.print(firstSetBit(s, n));
  }
}
 
// This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
  // Function to find
  // A bit is set or not
  static bool isSetBit(string s, int i)
  {
    return s[i] == '1';
  }
 
  // Function to find
  // First set bit
  static int firstSetBit(string s, int n)
  {
    int l = 0, r = n, m, ans = -1;
 
    // Applying binary search
    while (l <= r) {
 
      m = (l + r) / 2;
      if (isSetBit(s, m)) {
 
        // store the current
        // state of m in ans
        ans = m;
        r = m - 1;
      }
      else {
        l = m + 1;
      }
    }
 
    // Returning the position
    // of first set bit
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    string s = "111";
    int n = s.Length;
    Console.Write(firstSetBit(s, n));
  }
}
 
// This code is contributed by sanjoy_62.

Javascript

      // JavaScript code for the above approach
 
      // Function to find
      // A bit is set or not
      function isSetBit(s, i) {
          return s[i] == '1';
      }
 
      // Function to find
      // First set bit
      function firstSetBit(s, n) {
          let l = 0, r = n, m, ans = -1;
 
          // Applying binary search
          while (l <= r) {
 
              m = Math.floor((l + r) / 2);
              if (isSetBit(s, m)) {
 
                  // store the current
                  // state of m in ans
                  ans = m;
                  r = m - 1;
              }
              else {
                  l = m + 1;
              }
          }
 
          // Returning the position
          // of first set bit
          return ans;
      }
 
      // Driver code
      let s = "111";
      let n = s.length;
      document.write(firstSetBit(s, n));
 
     // This code is contributed by Potta Lokesh
  

Python

# Python Program to find Position
# Of leftmost set bit
 
# Function to find
# A bit is set or not
def isSetBit(s, i):
     
    return s[i] == '1'
 
# Function to find
# First set bit
def firstSetBit(s, n):
 
    l = 0
    r = n
    m = 0
    ans = -1
 
    # Applying binary search
    while (l <= r):
 
        m = (l + r) // 2
        if (isSetBit(s, m)):
 
            # store the current
            # state of m in ans
            ans = m
            r = m - 1
 
        else:
            l = m + 1
 
    # Returning the position
    # of first set bit
    return ans
 
# Driver code
 
s = "111"
n = len(s)
print(firstSetBit(s, n))
 
# This code is contributed by Samim Hossain Mondal.

输出：

0

时间复杂度： O(LogN)
辅助空间： o(1)