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  • 二分算法框架

      二分算法的前提是排序数组,通过折半查找的方式,将找到目标值的时间复杂度缩小到O(logn)

      最主要的3个问题

    • 查找某元素即返回索引
    • 查找某元素的左边界索引
    • 查找某元素的右边界索引

    二分算法框架:

    // 查找某元素即返回索引
    int binarySearch(int num[], int length, int target){
        if(length < 1)
            return -1;
        int left = 0;
        int right = length - 1;
        // 左右封闭
        while(left <= right){
            int middle = left + ((right - left) >> 1);
            if(num[middle] == target)
                return middle;
            else if(num[middle] > target)
                right = middle - 1;
            else if(num[middle] < target)
                left = middle + 1;
            // 全写else if,不写else,这样更好理解
        }
        // 找不到就直接返回
        return -1;
    }
    
    
    // 查找左边界
    int binaryLeftBound(int num[], int length, int target){
        if(length < 1)
            return -1;
        int left = 0;
        int right = length - 1;
        // 左右封闭法(注意也可以左闭右开)
        while(left <= right){
            int middle = left + ((right - left) >> 1);
            if(num[middle] == target)
                // 不着急返回,向左收缩范围
                right = middle - 1;
            else if(num[middle] > target)
                right = middle - 1;
            else if(num[middle] < target)
                left = middle + 1;
        }
        // left可能越界
        if(left >= length || num[left] != target)
            return -1;
        return left;
    }
    
    // 查找右边界
    int binaryRightBound(int num[], int length, int target){
        if(length < 1)
            return -1;
        int left = 0;
        int right = length - 1;
        // 左右封闭法(注意也可以左闭右开)
        while(left <= right){
            int middle = left + ((right - left) >> 1);
    
            if(num[middle] == target)
                // 不着急返回,向右收缩范围
                left = middle + 1;
            else if(num[middle] > target)
                right = middle - 1;
            else if(num[middle] < target)
                left = middle + 1;
        }
        if(right < 0 || num[right] != target)
            return -1;
        return right;
    }
    
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  • 原文地址:https://www.cnblogs.com/flyingrun/p/13569417.html
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