Binary search-2

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The title says that when a version is wrong , Later versions have errors , And try to reduce the sending of authentication requests as much as possible , This is obviously a test of dichotomy . Follow 1~100 Guess a number similar .

Ideas ：
First set the start and end values 0 and n. I thought of giving priority to the elimination of closeout , Because if the ending is the error source , The number of cycles is the highest , Will traverse to the innermost layer , Conduct whlie loop , Find the intermediate value , Determine whether the intermediate value is wrong , If it's wrong ,（ Then judge whether the previous one is wrong , If not , Is the error source , If it is , Set the starting value to the intermediate value ）, If the median is correct , Adjust the end value to the intermediate value + 1.

var solution = function(isBadVersion) {
    
    return function(n) {
    
        let start = 0
        let end = n
        if(isBadVersion(start)) return start 
        if(isBadVersion(end) && !isBadVersion(end - 1)) return end 
        while(start <= end) {
    
            let min = Math.floor((end - start) / 2) + start
            if(isBadVersion(min)) {
    
                if(!isBadVersion(min - 1)) {
    
                    return min
                }
                end = min
            }else {
    
                start = min + 1
            }
        }
    };
};

The code for the official answer seems to be more concise

var solution = function(isBadVersion) {
    
    return function(n) {
    
        let start = 0
        let end = n
        while(start < end) {
    
            let min = Math.floor((end - start) / 2) + start
            if(isBadVersion(min)) {
    
                end = min
            }else {
    
                start = min + 1
            }
        }
        return start
    };
};

First, set the start and end values , Start the traversal loop , Find the middle value , Narrow the range of start and end values according to the results of intermediate values . Until it is reduced to the same starting and ending , This value is the error source . It is worth noting that the judgment of traversal yes cannot be <=, If the start value and the end value are equal, the loop is entered , Will fall into a dead cycle .