LeetCode(一)求两个数之和( Two Sum)--------(OC版本)_oc 两数之和

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the sameelement twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
中文版:

给定一个整数数组和一个目标值，找出数组中和为目标值的两个数。

你可以假设每个输入只对应一种答案，且同样的元素不能被重复利用。

示例:

给定 nums = [2, 7, 11, 15], target = 9

因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
 

OC解

  //Two Sum

    //把数组元素和index生成对应的键值对,然后遍历数组元素若是sum - 元素 字典存在value则 value就是index

    NSArray *numArray = @[@(1),@(3),@(5),@(4),@(6)];

    NSInteger sum = 12;

    NSMutableDictionary *tempDict = [NSMutableDictionary dictionary];

    for (int i=0; i<[numArray count]; i++) {

        [tempDict setValue:@(i) forKey:[NSString stringWithFormat:@"%ld",[numArray[i] integerValue]]];

    }

    __block BOOL isHaveValueNumber = NO;

    [numArray enumerateObjectsUsingBlock:^(id  _Nonnull obj, NSUInteger idx, BOOL * _Nonnull stop) {

        NSString *keyStr = [NSString stringWithFormat:@"%ld",sum - [(NSNumber *)obj integerValue]];

        NSNumber* lastIndex  = tempDict[keyStr];

        if (lastIndex && idx < [lastIndex integerValue]) {

            NSLog(@"两个数值的index为%ld,%ld",[lastIndex integerValue],idx);

            isHaveValueNumber = YES;

        }

    }];

    if (!isHaveValueNumber) {

        NSLog(@"No vaild outputs");

    }